Projective Geometry

Lecture Notes

W. D. GILLAM

Boˇgazi¸ciUniversity 2014 Contents

Introduction iv

1 Transformations of the Plane 1 1.1 Linear transformations ...... 2 1.2 Isometries ...... 6 1.3 Orthogonal linear transformations ...... 7 1.4 Aﬃne transformations ...... 12 1.5 Classiﬁcation of isometries of the plane ...... 16 1.6 Algebraic automorphisms ...... 18 1.7 Exercises ...... 19

2 Parametric Plane Curves 20 2.1 Reparametrization ...... 21 2.2 Length and the arc length parametrization ...... 22 2.3 Curvature and the Frenet formulas ...... 23 2.4 Plane curves up to isometry ...... 25 2.5 Exercises ...... 26

3 Affine Algebraic Plane Curves 28 3.1 Polynomials and zero loci ...... 29 3.2 Smooth and singular points ...... 32 3.3 Change of variables and affine equivalence ...... 33 3.4 Five points determine a conic ...... 36 3.5 Plane conics up to affine equivalence ...... 38 3.6 Invariants of affine conics ...... 40 3.7 Exercises ...... 43

4 Projective Spaces 45 4.1 Motivation and construction ...... 45 4.2 Linear subspaces of projective space ...... 48 4.3 Real and complex projective spaces ...... 48 4.4 Projective transformations ...... 50

ii Contents iii

4.5 Exercises ...... 53

5 Projective Plane Curves 55 5.1 Homogeneous polynomials and zero loci ...... 56 5.2 Smooth and singular points ...... 57 5.3 Linear change of variables and projective equivalence ...... 58 5.4 Lines and linear functionals ...... 59 5.5 Invariants of aﬃne conics revisited ...... 61 5.6 Classiﬁcation of projective conics ...... 62 5.7 The cross ratio ...... 67 5.8 Intersection multiplicity ...... 69 5.9 The theorems of Pascal, Pappus, and Desargues ...... 70 5.10 Cubic curves and the group law ...... 72 5.11 Nodal and cuspital cubics ...... 76 5.12 Exercises ...... 78

6 Appendix 82 6.1 Groups and group actions ...... 82 6.2 Fields and division rings ...... 84 6.3 The Implicit Function Theorem ...... 86 6.4 Exercises ...... 87

Bibliography 88 Introduction

These notes are meant to serve as the text for an undergraduate course in elementary projective geometry. The current notes were written to accompany the course of the same title given at Boˇgazi¸ciUniversity in the fall of 2014. Some parts of these notes were recycled from notes I wrote to accompany a course called Fundamental Problems of Geometry, which I taught at Brown University in the spring of 2012. Both courses were intended for junior and senior mathematics majors. The second incarnation of this course was particularly aimed at students seeking a career in mathematics instruction. To this end, I made considerable effort to try to “survey” a variety of basic geometric material, with a focus on the geometry of the plane. In principle linear algebra is the only formal prerequisite, though there are many places where I find it convenient to use the language of group theory, though no actual group theory is really used—it is enough to know the definition of a group, a subgroup, and perhaps a normal subgroup. Also, the idea of a group acting on a set arises (at least implicitly) at many points, so it might be helpful to have an idea what this means—it is a completely elementary concept. Both times I taught the course, I considered devoting a lecture to the basic notions of group theory and group actions (orbits and stabilizers, in particular), but in the end I just mentioned the necessary concepts as they arose, or relegated them to the exercises—that seemed sufficient. At some point I may add an appendix to these notes covering this material. Let me make some remarks about “level of generality” and the like, mostly intended for the instructor. The usual issues about fields are skirted in these notes in much the same way they are skirted in a typical linear algebra class; that is, we pretty much think of a field as either R or C and we don’t much emphasize the difference even between these two fields, except that we can’t get away with this quite as long as we can in linear algebra because we will quickly consider non-linear equations like x2 + 1 = 0 which have radically different behaviour over these two fields. For the sake of exposition, I make most statements over R (instead of writing F or some such thing to denote a general field), but most of these are true over an arbitrary field, except where (I think) it should be reasonably clear from context that the particular field considered is of central importance to the statement of the result. In fact, I stubbornly insisted on using R as the “base field” throughout most of the notes, even though most of the “algebro-geometric” content of the notes (of which there is very little, by the way) would probably be easier and more

iv Introduction v

natural over C. 2 Although projective geometry and, in particular, the projective plane RP , are the main subject matter of these notes, a large part of the text is actually devoted to 2 various geometric considerations in the usual “affine” plane R . Without some of this “background” material, much of the projective geometry would seem unmotivated. I also wanted to emphasize the interplay of several different points of view on this subject matter coming from linear algebra, differential geometry, algebraic geometry, and classical axiomatic geometry. Chapter 1 is devoted to defining and studying various transformation groups, such as the group of invertible linear transformations, the group of isometries, and the group of affine transformations. The latter, in particular, is important because of its relationship with the group of projective transformations (the projective general linear group), which we discuss in the chapter on projective geometry. The material from this chapter is used throughout the rest of the text. In the first incarnation of the course, I didn’t say anything about isometries, but the group of isometries is so closely related to the group of affine transformations that it seemed strange to discuss one but not the other. Having discussed isometries in the Projective Geometry course, I couldn’t resist fleshing out this material to show how the group of isometries is used in a “real life” geometric study: To this end, I have included Chapter 2 on parametric curves in the plane. I thought it might be useful for pedagogical reasons to have the students return to a study of parametric curves, familiar from calculus courses, equipped with an understanding of the meaning of isometry and the group of isometries of the plane, as the idea of isometry is certainly lurking implicitly in the treatments of parametric curves one sees in any elementary calculus text! To make sure that this chapter contains some new and interesting content, I have included a proof of the (fairly simple) fact that two parametric curves in the plane are related by an orientation preserving isometry iff they have the same length and curvature. My treatment of this is taken from Do Carmo’s book Differential Geometry of Curves and Surfaces. The final two chapters consist of some elementary algebraic geometry of affine and n projective plane curves. We introduce the general projective space RP , but focus almost 2 exclusively on RP . We define the zero locus Z(f) of a polynomial f (or, rather, a homogeneous polynomial in the projective setting), and what it means to be a singular points of Z(f). We explain what it means for polynomials to be “affine equivalent” and for homogeneous polynomials to be “projectively equivalent.” To make this more concrete, we mention the classification of degree two polynomials in two variabes up to affine equivalence, though we only give a complete proof of the projective analog of this 2 classification. At the very end we give a brief study of cubic curves in RP , giving at least the rough idea of the group law on the set of smooth points of an irreducible cubic.

Chapter 1

Transformations of the Plane

2 2 Let us agree that a transformation (of the plane) is a bijective function f : R → R . Among all transformations we can single out various classes of transformations preserving 2 various additional structures possessed by R . In this chapter we will deﬁne and study the following types of transformations:

1. homeomorphisms

2. invertible linear transformations

3. isometries

4. orthogonal linear transformations

5. aﬃne transformations

6. algebraic automorphisms

n One can consider these types of transformations more generally in the case of R , or n over other ﬁelds. Instead of R , one could work with an abstract vector space, or with a vector space equipped with some extra structure. We will make some remarks about 2 these more general settings, but our main focus will be on the case R , as this will be most important in the later chapters. The set of transformations of each type forms a group. In some cases, we will say something about the structure of this group. Although detailed deﬁnitions will be given in the sections of this chapter, we will give a summary now for the convenience of the reader. A homeomorphism is a continuous 2 2 −1 2 2 bijection f : R → R with continuous inverse f : R → R . Homeomorphisms are the most general type of transformation that will be of any use to us—every other type of transformation we consider will be a homeomorphism. Homeomorphisms preserve 2 properties of subsets of R such as connectedness, being open, or being compact (closed and bounded). 2 2 An invertible linear transformation is a bijection f : R → R that “commutes with scalar multiplication and vector addition.” The importance of invertible linear 2 transformations results mainly from their usage in linear algebra. An isometry of R is a 2 2 bijection f : R → R that preserves distance. Isometries can be useful when studying

1 2 Chapter 1 Transformations of the Plane

2 various geometric properties of subsets of R such as lengths, areas, angles, and so forth. 2 2 One can define an orthogonal linear transformation to be a map f : R → R which is both an isometry and a linear transformation. We will describe all of these. An 2 2 affine transformation is a map f : R → R expressible as a composition of a linear transformation and a translation. Affine transformations arise naturally when studying 2 more “algebraic” subsets of R (zero loci of polynomials) and their properties. They are also closely related to the projective transformations (§4.4) that we will study when we introduce projective geometry in Chapter 4. An algebraic automorphism is a bijection 2 2 f : R → R such that the coordinates of both f and its inverse are given by polynomial functions of two variables. These are the most subtle kind of transformation that we will consider—we won’t have too much to say about them, but it would be amiss not to mention these along with the others! One reason to be interested in these sorts of transformations is the following: If one is 2 interested in calculating some quantity Q(A) associated to a subset A ⊆ R (or perhaps 2 to several subsets A1,...,An ⊆ R ), it is often convenient to calculate the corresponding quantity Q(f(A)) for f(A) for some transformation f for which it is known a priori that Q(A) = Q(f(A)). Of course this seems silly when expressed in such abstract terms, 2 but imagine, for example, the problem of computing the circumference of a circle in R , 2 or, more generally, the length `(A) of some other curve A in R . Because this problem involves doing significant computation (integration), it is usually helpful to move A to some other position via an isometry f (for example: move the circle so it is centered at the origin), which won’t effect the result of the computation (because one knows “a priori” that length is isometry invariant), but which will make the computation easier to actually carry out. Similar considerations are common in linear algebra, where various operations are naturally defined and studied in one basis; one then uses invertible linear transformations to re-express things in the standard basis. We will make use of the various kinds of transformations for purposes such as this in the other chapters of the book.

1.1 Linear transformations

Although I assume the student has some background in linear algebra, let me briefly review some of the more relevant material. Throughout, we work with a fixed field K, often called the base field. The reader unfamiliar with the general notion of a field should simply focus on the case where K is the field Q of rational numbers, the field R of real numbers, or the field C of complex numbers. Let

n K := {(x1, . . . , xn): x1, . . . , xn ∈ K} be the set of ordered n-tuples of elements of K. We will use symbols like x, y, etc. to n denote elements of K , writing x1, . . . , xn for the coordinates of x. n From the point of view of linear algebra, the set K is equipped with two important n structures: (vector) addition and scalar multiplication. For x, y ∈ K , we let

x + y := (x1 + y1, . . . , xn + yn) 1.1 Linear transformations 3

n be the “vector sum” of x and y. Given λ ∈ K and x ∈ K , we set

λx := (λx1, . . . , λxn). The vector λx is called a scalar multiple of x; the term rescaling of x is reserved for a ∗ vector of the form λx with λ ∈ K := K \{0}. Vector addition and scalar multiplication satisfy various properties, which one is led to investigate and “axiomatize”: for example, scalar multiplication “distributes over vector addition.” After thinking about these properties, one is eventually led to deﬁne a vector space (or a vector space over K if K is not clear from context) to be a set V equipped with two functions, written

V × V → V (u, v) 7→ u + v R × V → V (λ, v) 7→ λv, and called, respectively, (vector) addition and scalar multiplication satisfying various axioms that we will not repeat here (see any linear algebra textbook). n It is customary in linear algebra to write the elements x of K as column vectors, but in most of the inline text of these notes I will usually write these as row vectors x = (x1, . . . , xn) for aesthetic reasons. I could put some kind of “transpose” notation in to be precise, but I think it will always be clear when I am using row vectors or column vectors. I usually use u, v, etc. for elements of an abstract vector space, which I usually denote U, V , or W . In linear algebraic context, I try to use Greek letters like λ, µ, and so forth for elements of K, also called scalars. Deﬁnition 1.1.1. If V and W are vector spaces, a linear transformation from V to W is a function f : V → W satisfying f(λv) = λf(v) and f(u + v) = f(u) + f(v) for every λ ∈ K, u, v ∈ V . Informally, one says that f “respects addition and scalar multiplication.” Let HomK(V,W ) be the set of linear transformations f : V → W . It is easy to check that a composition of linear transformations is a linear transformation, so that usual composition of functions deﬁnes a “composition of linear transformations”

HomK(V,W ) × HomK(U, V ) → HomK(U, W ) (f, g) 7→ fg for vector spaces U, V, W . It is also easy to see that, if f : V → W is a bijective linear transformation, then its inverse f −1 : W → V is also a (bijective) linear transformation.

Deﬁnition 1.1.2. An m × n matrix A = (Ai,j) consists of elements Ai,j ∈ K, called the entries of A, one for each

(i, j) ∈ {1, . . . , m} × {1, . . . , n}.

We write Mat(m × n) for the set of m × n matrices; if there is any ambiguity about K, then we write Mat(m × n, K). 4 Chapter 1 Transformations of the Plane

One thinks of the entries Ai,j of a matrix A as being arranged in a rectangular array such that Ai,j is in row i and column j. Thus, for example, a 2 × 3 matrix A is written in the form ! A A A A = 1,1 1,2 1,3 . A2,1 A2,2 A2,3

It is common to drop the comma in the subscript, writing Aij instead of Ai,j, though this is a bit sloppy since “ij” might be mistaken for the product of i and j. The entries of a 2 × 2 matrix are almost always called a, b, c, d rather than A1,1, A1,2, A2,1, and A2,2, so that a typical 2 × 2 matrix is written ! a b A = . (1.1) c d

The fundamental operation with matrices is the (matrix) product: Given an m × n matrix A and an n × k matrix B, one deﬁnes an m × k “product” matrix AB by setting

n X (AB)i,j := Ai,pBp,j. p=1

To each square matrix A ∈ Mat(n × n), one can associate a real number det A, called the determinant of A. If A is 1 × 1, one not surprisingly has det A = A1,1 equal to the unique entry of A. If A is a 2 × 2 matrix as in (1.1), then det A = ad − bc. In general, in an elementary linear algebra class, one deﬁnes det A by some kind of inductive procedure (“expansion by minors”), though there are more satisfying ways of doing this. In any case, the most important property of the determinant is that

det(AB) = (det A)(det B) for all n × n matrices A, B. The “matrix product” is motivated by linear algebraic considerations as follows: n th For i ∈ {1, . . . , n}, let ei ∈ K be the point with i coordinate 1 and all other 2 n m coordinates zero. For example, e1 = (1, 0) and e2 = (0, 1) in K . If f : K → K is th a linear transformation, we deﬁne an m × n matrix A = Af by letting the i column n (i = 1, . . . , n) of A be f(ei) ∈ K (viewing this as a column vector). For example, if 2 2 f : R → R and f(e1) = (a, c), f(e2) = (b, d), then ! a b Af = . c d

n m Given an m × n matrix A, we deﬁne a function fA : K → K by letting fA(x) be the matrix product Ax deﬁned above, viewing x as a column vector (n × 1 matrix). The following standard result of linear algebra says that “matrices are the same thing as linear transformations and matrix multiplication is the same thing as composition:” 1.1 Linear transformations 5

Proposition 1.1.3. The maps f 7→ Af and A 7→ fA deﬁned above are inverse bijections n m between HomK(K , K ) and Mat(m × n). Under these bijections, matrix multiplication

Mat(m × n) × Mat(n × k) → Mat(m × k) corresponds to composition of linear transformations

n m k n k n HomK(K , K ) × HomK(K , K ) → HomK(K , K ).

That is, fAB = fAfB and Afg = Af Ag.

Recall the following standard result of linear algebra, which, among other things, explains the importance of the determinant:

n n Proposition 1.1.4. For a linear transformation f : K → K , the following are equivalent:

n 1. f(e1), . . . , f(en) is a basis for K .

2. f is surjective.

3. f is injective.

4. f is bijective.

5. Ker f = {0}.

n n 6. There is a linear transformation g : K → K such that gf(x) = fg(x) = x for all n x ∈ R .

7. The determinant det Af of the matrix Af associated to f is non-zero.

n n Deﬁnition 1.1.5. A linear transformation f : K → K satisfying the equivalent conditions of the above proposition is called an invertible linear transformation.

It is clear that a composition of invertible linear transformations is an invertible linear n n transformation. The set of invertible linear transformations f : K → K forms a group under composition, denoted GLn(K), and often called the general linear group. Blurring the distinction between a linear transformation and the corresponding matrix, an element of GL2(K) can be viewed as a 2 × 2 matrix ! a b A = c d where det A = ad − bc 6= 0. 6 Chapter 1 Transformations of the Plane

1.2 Isometries

Recall that a metric d on a set X is a function

d : X × X → R satisfying the properties: 1. d(x, y) ≥ 0 for all x, y ∈ X with equality iﬀ x = y. 2. d(x, y) = d(y, x) for all x, y ∈ X. 3. d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X. A metric space (X, d) is a set X equipped with a metric d. We usually just write X to denote a metric space, leaving d implicit; we often write dX for the metric on X if there is any chance of confusion (for example, if we are considering more than one metric space). For example, it is not hard to check that

n n n d : R × R → R v u n uX 2 d(x, y) := t (xi − yi) i=1 n n is a metric on R , called the standard metric. By default, we regard R as a metric space by using the standard metric unless some other metric is explicitly mentioned. In general, if d is a metric on a set X and A is a subset of X, then by restriction d determines a metric on A, also abusively denoted d. In particular, the standard metric n n on R determines a metric on any subset A ⊆ R ; we always use this metric to view A as a metric space, unless explicitly mentioned to the contrary. One defines continuity for maps between metric spaces by using the familiar “epsilon- delta definition:” Definition 1.2.1. A function f : X → Y between metric spaces X,Y is called continuous iff for every x ∈ X and every positive real number , there is a positive real number 0 0 0 δ = δ(x, ) such that dY (f(x), f(x )) < whenever x is a point of X with dX (x, x ) < δ. A homeomorphism f : X → Y is a continuous bijection whose inverse f −1 : Y → X is also continuous. It is easy to see that a composition of continuous maps between metric spaces is continuous, hence a composition of homeomorphisms is a homeomorphism. In particular, n n n the set Homeo(R ) of homeomorphisms f : R → R is a group under composition. It n n is also not hard to check that any linear transformation f : R → R is continuous. It follows that the group GLn(R) of invertible linear transformations is a subgroup of n Homeo(R ). Definition 1.2.2. A function f : X → Y between metric spaces X and Y is called 0 0 0 distance preserving iff dX (x, x ) = dY (f(x), f(x )) for all x, x ∈ X. An isometry f : X → Y is a distance preserving bijection. 1.3 Orthogonal linear transformations 7

The following facts are readily established:

1. A composition of distance preserving functions is distance preserving.

2. A distance preserving function is continuous (one can take δ = ) and one-to-one (injective).

3. The inverse of an isometry is an isometry.

4. Every isometry is a homeomorphism.

5. A composition of isometries is an isometry.

n n n In particular, the set Isom(R ) of isometries f : R → R is a subgroup of the group n Homeo(R ). The deﬁnitions above yield notions of continuity, isometry, etc. for functions between n 2 2 subsets of R . We will see later that any distance preserving function R → R is n automatically bijective (i.e. is an isometry). (This is true, more generally, for R .) 2 Indeed, we will completely classify the isometries of R , and this will fall out as a consequence—the reader may wish to try to directly prove this assertion.

n Example 1.2.3. (Translations) Let K be a ﬁeld. Fix t ∈ K . Deﬁne a function n n n n ft : K → K by ft(x) := t + x. When K = R, the function ft : R → R preserves distance because v u n uX 2 d(ft(x), ft(y)) = t ((ti + xi) − (ti + yi)) i=1 v u n uX 2 = t (xi − yi) i=1 = d(x, y).

n n n A function f : K → K equal to ft for some t ∈ K is called a translation—the t will be unique since we can recover t from ft by the obvious formula t = ft(0). Notice that the composition fsft of two translations is the translation fs+t. The set of translations n n n K → K is a group under composition of functions; this group is isomorphic to K (under vector addition) via the maps t 7→ ft, f 7→ f(0). The inverse of the translation ft n n is the translation f−t. In particular, every translation ft : R → R is an isometry.

1.3 Orthogonal linear transformations

Given a real vector space V (vector space over the ﬁeld of real numbers R), recall from linear algebra that an inner product (more precisely: positive deﬁnite inner product) on V is a function V × V → R, usually written (u, v) 7→ hu, vi, satisfying the conditions 1. hu, vi = hv, ui 8 Chapter 1 Transformations of the Plane

2. hu, ui ≥ 0 with equality iﬀ u = 0

3. hu + u0, vi = hu, vi + hu0, vi

4. hλu, vi = λhu, vi

0 for all u, u , v ∈ V , λ ∈ R. For example,

n n h , i : R × R → R n X hx, yi := xiyi i=1 n deﬁnes an inner product on R , called the standard inner product. An inner product space is a real vector space V equipped with an inner product. An inner product h , i on V gives rise to a metric d on V deﬁned by

d(u, v) := hu − v, u − vi.

Two vectors u, v in an inner product space V are called orthogonal iﬀ hu, vi = 0. More generally, one deﬁnes the angle θ ∈ [0, π] between two non-zero vectors u, v in V by the formula hu, vi cos θ = (1.2) |u||v| where the symbol

|u| := d(u, 0) = phu, ui denotes the magnitude of u. (One checks that the right hand side of (1.2) is in [−1, 1] so that this makes sense.) n Unless mentioned to the contrary, we always regard R as an inner product space by n equipping it with the standard inner product. The standard inner product on R gives n rise to the standard metric on R . Let us check that the notion of the angle between two n vectors deﬁned by the standard inner product is the usual angle in R . Recall that the Law Of Cosines says that in any triangle with sides of length a, b, c, we have

c2 = a2 + b2 − 2ab cos C, where C is the angle opposite the side of length c. If we apply this to the “triangle” n whose vertices are 0, and two non-zero vectors x, y ∈ R , letting c be the side connecting x to y, we ﬁnd

hu, vi = |u|2 + |v|2 − 2|u||v| cos θ.

If we expand out the inner product on the left, cancel the |u|2 + |v|2 appearing on both sides and divide by −2, we ﬁnd that the angle θ between u and v is the same as the 1.3 Orthogonal linear transformations 9

n “abstractly defined” angle θ. In particular, x, y ∈ R are orthogonal in the usual sense iff hx, yi = 0. One says that a list of n vectors v1, . . . , vn in an n-dimensional inner product space V is an orthonormal basis for V iff hvi, vji = δi,j is zero when i 6= j and one when i = j for all i, j ∈ {1, . . . , n}. An orthonormal basis is, in particular, a basis: By linear algebra, it suffices to check that the vectors v1, . . . , vn are linearly independent. Indeed, if we Pn have i=1 λivi = 0, then taking h , vji and using properties of the inner product and the orthogonality assumption on the vi, we see that λj = 0. n The standard inner product on R gives rise to a nice relationship between the linear n n algebra of R and the isometries of R :

n n Proposition 1.3.1. For a linear transformation f : R → R , the following are equivalent:

n 1. hf(x), f(y)i = hx, yi for all x, y ∈ R .

2. The columns of the matrix A = Af associated to f form an orthonormal basis for n R .

3. f preserves distance

n 4. f is an isometry of R .

th Proof. To see that (1) implies (2), just note that the i column of Af is f(ei), so we have to show that hf(ei), f(ej)i = δi,j. But by (1) we have

hf(ei), f(ej)i = hei, eji

n and it is clear from the deﬁnition of the standard inner product on R that this is δi,j. Pn Pn To see that (2) implies (1), write x = i=1 xiei, y = i=1 yiei for all real numbers Pn Pn x1, . . . , xn and y1, . . . , yn. Then f(x) = i=1 xif(ei) = i=1 Ai, where Ai = f(ei) is the th i column of A = Af . Using this, the analogous formula for f(y), the hypothesis on A, and properties of h , i, we compute

* n n + X X hf(x), f(y)i = xiAi, yiAi i=1 i=1 n X = xiyjhAi,Aji i,j=1 n X = xiyjδi,j i,j=1 n X = xiyj i=1 = hx, yi. 10 Chapter 1 Transformations of the Plane

n To see that (1) implies (3), we compute, for any x, y ∈ R that

d(f(x), f(y)) = hf(x) − f(y), f(x) − f(y)i = hf(x − y), f(x − y)i = hx − y, x − yi = d(x, y) using linearity of f for the second equality and (1) for the third equality. th To see that (3) implies (2), let Ai = f(ei) denote the i column of A = Af . First note that

hAi,Aii = hAi − 0,Ai − 0i

= d(Ai, 0)

= d(f(ei), f(0))

= d(ei, 0), so we see that each column of A has magnitude one. We still have to show that any two distinct columns Ai, Aj of A are orthogonal, so ﬁx distinct i, j ∈ {1, . . . , n} and let θ be the angle between Ai and Aj. Since f preserves distance, we have

d(Ai,Aj) = d(f(ei), f(ej))

= d(ei, ej) √ = 2.

Now the Law Of Cosines gives

2 2 = d(Ai,Aj) 2 2 = |Ai| + |Aj| − 2|Ai||Aj| cos θ = 2 − 2 cos θ, so we must have cos θ = 0, hence Ai and Aj are orthogonal, as desired. n To see that (3) implies (4), we just need to show that a linear transformation f : R → n R which preserves distance is surjective. (Recall that every distance preserving map is injective and that an isometry is a distance preserving bijection.) Since f is injective, we can just use the general linear algebra fact that an injective linear transformation between vector spaces of the same (ﬁnite) dimension is bijective. Alternatively, we could use the fact that (3) implies (2) to establish bijectivity of f. Obviously (4) implies (3).

n n Definition 1.3.2. A linear transformation f : R → R satisfying the equivalent conditions of Proposition 1.3.1 is called an orthogonal linear transformation, or simply an orthogonal transformation. Similarly, a matrix A ∈ Mat(n×n) is called an orthogonal matrix iff the corresponding linear transformation fA is an orthogonal linear transformation n (iff the columns of A form an orthonormal basis for R ). 1.3 Orthogonal linear transformations 11

Since a composition of linear transformations is a linear transformation and a composition of isometries is an isometry, we see that a composition of orthogonal linear transformations is an orthogonal linear transformation. The group of orthogonal linear n n transformations R → R is denoted On. The group On may be viewed as a subgroup of n both GLn(R) and Isom(R ). Indeed, the proposition above shows that n On = GLn(R) ∩ Isom(R ),

n viewing all of these groups as subgroups of Homeo(R ). n The orthogonal transformations of R can be explicitly described: 2 2 Proposition 1.3.3. Suppose f : R → R is an orthogonal transformation, with associated matrix A = Af . Then the determinant of A is ±1. If det A = 1, then there is a unique angle θ ∈ R/2πZ such that ! cos θ − sin θ A = . sin θ cos θ

If det A = −1, then there is a unique angle θ ∈ R/2πZ such that ! cos θ sin θ A = . sin θ − cos θ

Proof. If f is an orthogonal transformation, then f(e1) and f(e2) (the columns of A) have magnitude one, hence lie on the unit circle

1 2 2 2 S := {(x, y) ∈ R : x + y = 1}, so we can write f(e1) = (cos θ, sin θ) for a unique θ ∈ R/2πZ. Since f(e2) must also be a point of the unit circle orthogonal to f(e1), the second column of A is either

(cos(θ + π/2), sin(θ + π/2)) = (− sin θ, cos θ) or

(cos(θ − π/2), sin(θ − π/2)) = (sin θ, − cos θ).

2 Remark 1.3.4. Proposition 1.3.3 characterizing the orthogonal transformations of R in particular shows that each such orthogonal transformation has determinant ±1. It is n true more generally that any orthogonal transformation of R has determinant ±1. To see this, observe that for any square matrix A with real entries, we have

tr (A A)ij = hAi,Aji,

tr th where A denotes the transpose of A and Ai denotes the i column of A. In particular, we see that A is an orthogonal matrix iﬀ AtrA = Id. Since det A = det Atr by general theory of determinants, this shows that (det A)2 = det Id = 1. 12 Chapter 1 Transformations of the Plane

2 2 Example 1.3.5. (Rotations) If f : R → R is an orthogonal transformation with det Af = 1 (also called a special orthogonal transformation) and we write f as in the above proposition, then, geometrically, f is given by counter-clockwise rotation around the 2 origin through the angle θ. Indeed, one checks readily that d(0, f(x) = d(0, x) for x ∈ R , 2 and that if η is the counter-clockwise angle between the positive x-axis and x ∈ R \{0}, 2 then the counter-clockwise angle between the positive x-axis and f(x) ∈ R \{0} is η + θ. The group SO2 of special orthogonal transformations is a normal subgroup of O2. One checks easily that the map taking θ ∈ R/2πZ to the first matrix A = Aθ in the above proposition (or rather, to the corresponding orthogonal transformation fA) defines an isomorphism of groups ∼ (R/2πZ, +) = SO2 . 2 2 One can also define an isometry f : R → R by rotating around an arbitrary point of 2 R . The reader can check that such an isometry can be expressed as a composition of translations (Example 1.2.3) and a rotation around the origin. Example 1.3.6. (Reflections) The matrix ! 1 0 A := 0 −1

2 2 is clearly an orthogonal matrix, so the associated linear transformation f = fA : R → R is an orthogonal linear transformation, hence, in particular, an isometry. Note that f(x, y) = (x, −y), so that f can be described geometrically by reﬂecting over the x-axis. Since ! ! ! cos θ sin θ cos θ − sin θ 1 0 = , sin θ − cos θ sin θ cos θ 0 −1

2 the classification of orthogonal transformations of R in Proposition 1.3.3 shows that every such transformation is either a rotation around the origin, or the reflection across the x-axis followed by such a rotation. 2 2 More generally, one can define an isometry rL : R → R by reflecting across an 2 arbitrary line L ⊆ R . The reader can check that such an isometry can be expressed as a composition of rotations, translations, and the reflection over the x-axis discussed previously. In particular, it is a standard linear algebra exercise to show that reflection over a line passing through the origin is an orthogonal linear transformation of determinant −1.

1.4 Aﬃne transformations

n Let K be a field, K the set of ordered n-tuples of elements of K, as in §1.1. n n Definition 1.4.1. An affine transformation is a function f : K → K expressible as the composition of a linear transformation (Definition 1.1.1) followed by a translation (Example 1.2.3). 1.4 Affine transformations 13

The first thing to observe is that the expression of an affine transformation f = ftA as a linear transformation A followed by a translation ft is unique because we can recover t from f via f(0) = ft(A(0)) = ft(0) = t and then we can recover A from f and t by the obvious formula A = f−tf. In other words, n n n if we fix A ∈ GLn(K) and t ∈ K , then we get an affine transformation [A, t]: K → K by setting [A, t](x) := Ax + t. Every affine transformation is of the form [A, t] for a n unique A ∈ GLn(K) and t ∈ K , so that the set of affine transformations is in bijective n correspondence with the set GLn(K) × K . n Affine transformations form a group under composition, denoted Aff(K ). It is n important to understand that, although Aff(K ) is in bijective correspondence with n GLn(K) × K via the bijection described above, this bijection is not an isomorphism n n of groups between Aff(K ) and the product group GLn(K) × K . To see why this is the case, let us calculate the composition of two affine transformations [A, t] and [A0, t0]. n Given x ∈ K , we compute ([A, t] ◦ [A0, t0])(x) = [A, t](A0x + t0) = A(A0x + t0) + t = (AA0)x + (At0 + t), so the composition of affine transformations is given by [A, t][A0, t0] = [AA0, At0 + t]. (1.3) It follows from (1.3) that the inverse of the affine transformation [A, t] is given by [A, t]−1 = [A−1, −A−1t]. (1.4)

n n If we view Aff(K ) as the set of pairs [A, t] ∈ GLn(K) × K with composition law (1.3), n n then the group of translations K is identified with the subgroup of Aff(K ) consisting of the pairs [A, t] where A = Id, and the group GLn(K) of invertible linear transformations n is identified with the subgroup of Aff(K ) consisting of the pairs [A, t] where t = 0. n n Theorem 1.4.2. The group of translations (K , +) is a normal subgroup of Aff(K ) and n every element of Aff(K ) can be written uniquely as a composition of a translation and n an invertible linear transformation. In other words, Aff(K ) is a semi-direct product of n (K , +) and GLn(K). n Proof. The only statement not proved in the discussion above is the normality of (K , +). We calculate the conjugate of a translation [Id, s] by an arbitrary affine transformation [A, t] by using the formulas (1.3) and (1.4): [A, t][Id, s][A, t]−1 = [A, t][Id, s][A−1, −A−1t] = [A, As + t][A−1, −A−1t] = [Id,A(−A−1t) + As + t] = [Id, As]. This is a translation, as desired. 14 Chapter 1 Transformations of the Plane

n At present, the consideration of the groups Aﬀ(K ) may seem somewhat unmotivated. In fact, these groups are perhaps the most important groups that arise in geometry—here are some reasons why:

1. They are used to define and study (affine) linear changes of coordinates and (affine) linear changes of variables, hence they play a role in problems of classifying polynomials and (affine) plane curves.

n 2. The group Aﬀ(K ) is closely related to the projective general linear group, and n n n hence to the geometry of the projective space KP and the inclusion K ,→ KP , which we shall study later.

2 3. As we shall see in §1.5, the group of affine transformations Aff(R ) is closely related 2 to the group of isometries of R . 4. We will see in Theorem 1.4.5 that affine transformations arise naturally when 2 considering the geometry of lines in R . n n Proposition 1.4.3. Suppose P0,...,Pn ∈ K are (n+1) points of K in general position, n meaning they are not all contained in a translate of some linear subspace V ⊆ K of n dimension < n. Then there is a unique affine transformation [A, t] ∈ Aff(K ) such that [A, t](0) = P0 and [A, t](ei) = Pi for i = 1, . . . , n.

2 Remark 1.4.4. When n = 2, the condition in Proposition 1.4.3 that P, Q, R ∈ K be in “general position” just says that P, Q, R are not collinear (are not contained in a line).

Proof. Since the Pi are in general position, the vectors P1 − P0,...,Pn − P0 must be n linearly independent, hence by linear algebra, they form a basis for K and there is a n n unique linear transformation A : K → K with Aei = P1 −P0. The aﬃne transformation [A, P0] given by the linear transformation A followed by translation along P0 is clearly the unique aﬃne tranformation with the desired properties.

2 2 Theorem 1.4.5. A function f : R → R is an affine transformation iff f is a continuous 2 function with the following property (*): For any P, Q, R ∈ R , P, Q, R are collinear iff f(P ), f(Q), f(R) are collinear.

2 Proof. It is clear that any affine transformation of R is a continuous function satisfying (*) and that a composition of continuous functions satisfying (*) is also a continuous function satisfying (*). Suppose now that h is a continuous function satisfying (*). Then h(0), h(e1) and h(e2) are non-collinear, so by the previous proposition there is an affine 2 2 transformation g : R → R with g(0) = h(0), g(e1) = h(e1) and g(e2) = h(e2). We claim that h = g—equivalently, g−1h = Id. Since g−1 is an affine transformation, our earlier −1 observations show that f := g h is a continuous function satisfying (*) and fixing 0, e1, and e2. We thus reduce to proving that such a function f must be the identity. 2 2 In the rest of the proof we assume that f : R → R satisfies (*). We first note that 2 f must be injective: Indeed, suppose f(P ) = f(Q), but P 6= Q. Pick any point R ∈ R so that P, Q, R are not collinear. But f(R), f(P ), and f(Q) are certainly collinear, 2 contradicting (*). We next note that if P,Q are distinct points of R , then the image 1.4 Affine transformations 15 f(PQ) of the line PQ must be contained in the line f(P )f(Q). Finally, we note that if PQ, RS are parallel lines, then the lines f(P )f(Q) and f(R)f(S) must be parallel. From these simple remarks, we next deduce the key observation: If f fixes three vertices P, Q, R of a parallelogram with vertices P, Q, R, S, then it must also fix the forth vertex S. To see this, we can assume the vertices are labelled so that the lines PQ and RS are parallel and the lines PS and QR are parallel. Since f fixes P and Q and satisfies (*), it takes the line PQ into itself, so, since f also fixes R, it takes RS into Rf(S), hence, since f takes parallel lines to parallel lines, f(S) must lie on the line RS. But we see similarly that f(S) must also lie on the line PS, so f must fix S. Now, if f fixes 0, e1, e2, then we can repeatedly apply the above observation to parallelograms of one of the following five forms

1. P , P + e1, P + e2, P + e1 + e2

2. P , P + e1, P + e1 + e2, P + 2e1 + e2

3. P , P + e1, P + e2, P − e1 − e2

4. P , P + e1, P + e2, P + e1 − e2

5. P , P + e1, P − e2, P + e1 + e2

2 2 2 2 with P ∈ Z ⊆ R chosen appropriately to see that f must fix each point of Z ⊆ R . For example: We first apply the observation to the parallelogram of the first type with P = 0 to see that f fixes e1 + e2. Next we repeatedly apply the observation to parallelograms of the second and third types to see that f fixes all points of the form (a, b) with a ∈ Z, b ∈ {0, 1}. Finally we apply the observation repeatedly to parallelograms of the fourth 2 and fifth types to conclude that f fixes every point of Z . Next we argue that if f fixes 0, e1, and e2, then it must fix all points with rational 2 2 2 coordinates. Consider a point S = (x, y) ∈ Q ⊆ R . Take any point P ∈ Z not equal 2 to S. Since the line PS has rational slope and contains P ∈ Z , it will contain some 2 point R ∈ Z distinct from P . Since f fixes P and R it takes PR into itself, hence 0 2 f(S) ∈ PR = PS. Now just take another point P ∈ Z not lying on the line PR = PS and repeat the same argument to show that f(S) ∈ P 0S, hence f(S) = S as desired. 2 2 Now we have shown that a function f : R → R satisfying (*) and fixing 0, e1, e2 fixes all points with rational coordinates, hence, if f is also continuous, it must be the identity.

2 2 Remark 1.4.6. For a general field K, a function f : K → K satisfying (*) in Theo- 2 2 rem 1.4.5 is called a collineation. (One defines a line in K to be a subset of K arising 2 as a translation of a one-dimensional linear subspace of the vector space K .) Here is one way to construct collineations: Let G be the automorphism group of the field K 2 2 (the “Galois group” of K). Each σ ∈ G gives a collineation fσ : K → K by applying σ to the coordinates; let us call such a collineation a Galois collineation. Notice that each Galois collineation fixes (0, 0), (1, 0), and (0, 1). In fact, it can be shown that any 2 collineation of K fixing (0, 0), (1, 0), and (0, 1) is a Galois collineation. (The argument for this is elementary—I have omitted it only because it is a bit long and tedious.) It 16 Chapter 1 Transformations of the Plane

2 follows from the argument in the above proof that any collineation of K can be written 2 as a composition of an affine transformation of K and a Galois collineation. It can also be shown that the automorphism group of the field R is trivial (the idea here is that the ordering of R can be interpreted in terms of the field structure because, in R, being ≥ 0 is the same thing as being a square, which is a field-theoretic notion). It follows that every 2 collineation of R is, in fact, an affine transformation. In other words, Theorem 1.4.5 remains true even if the word “continuous” is deleted. See Propositions 3.11 and 3.12 in [H2].

2 2 Remark 1.4.7. The proof of Theorem 1.4.5 shows that any collineation f : Q → Q is an aﬃne transformation.

1.5 Classiﬁcation of isometries of the plane

2 In this section we will classify all isometries of R . 2 Lemma 1.5.1. Let C1, C2 be circles in R with centers P1,P2, respectively. Assume that C1 and C2 intersect in precisely two points: P and Q. (In particular this implies that P1 =6 P2.) Then the line segment PQ is orthogonal to the line P1P2 and PQ intersects P1P2 at the midpoint of PQ.

2 2 2 Proof. Let f : R → R be the isometry of R given by reﬂection (Example 1.3.6) across the line L = P1P2. Since P1,P2 ∈ L, we have f(Pi) = Pi (i = 1, 2). Since f is an isometry with f(Pi) = Pi, f takes Ci into itself, hence f(P ), f(Q) are in C1 ∩ C2 = {P,Q}. We cannot have f(P ) = P and f(Q) = Q, for then P , Q, and Pi would all lie on L (the ﬁxed locus of f), and P,Q would be equidistant from both P1 and P2, hence we would have P = Q. So, since f is bijective, we must have f(P ) = Q and hence P = f(f(P )) = f(Q). The conclusion follows easily.

2 2 Lemma 1.5.2. Suppose f : R → R is a distance preserving function such that 2 f(P1) = P , f(P2) = P2 and f(P3) = P3 for three non-collinear points P1,P2,P3 ∈ R . Then f is the identity.

2 Proof. We need to show that f(Q) = Q for an arbitrary Q ∈ R . Suppose—toward a 2 contradiction—that f(Q) 6= Q for some Q ∈ R . Then Q cannot be one of the Pi, so the distances di := d(Pi,Q) are positive. Let Ci be the circle of radius di centered at Pi. Then Q ∈ C1 ∩ C2 ∩ C3. Since f preserves distance and ﬁxes Pi it takes Ci into Ci, hence f(Q) ∈ C1 ∩ C2 ∩ C3. Suppose i, j ∈ {1, 2, 3} are distinct. Then the circles Ci ∩ Cj cannot intersect in more than two points, for then they’d be equal and we’d have Pi = Pj, contradicting the assumption that the Pi aren’t collinear. We conclude that Ci ∩ Cj = {Q, f(Q)}. Let Lij be the line through Pi and Pj. By the previous lemma, the lines L12, L13, and L23 are all orthogonal to Qf(Q) and each intersects Qf(Q) at the midpoint of the line segment Qf(Q), so we have L12 = L13 = L23, contradicting the assumption that the Pi are not collinear.

2 2 Lemma 1.5.3. A function f : R → R is distance preserving and satisfies f(0) = 0 iff f is an orthogonal linear transformation. 1.5 Classification of isometries of the plane 17

Proof. Since any linear transformation fixes zero (takes 0 to 0), Proposition 1.3.1 implies that any orthogonal linear transformation is an isometry fixing zero. Now suppose 2 2 f : R → R is distance preserving and satisfies f(0) = 0. Then, since f preserves distance, f(e1) and f(e2) have to be on the unit circle (i.e. at distance one from the origin) and √ d(f(e1), f(e2)) = d(e1, e2) = 2.

As in the proof of Proposition 1.3.1, the Law Of Cosines then implies that f(e1) and f(e2) are orthogonal unit vectors, so the matrix A with columns f(e1), f(e2) is an orthogonal 2 2 matrix and the corresponding linear transformation fA : R → R is an orthogonal linear transformation with fA(e1) = f(e1), fA(e2) = f(e2), and fA(0) = 0 = f(A). Then the −1 2 composition fA f is an isometry of R ﬁxing the three non-collinear points 0, e1, e2, hence −1 fA f = Id by the previous lemma, hence f = fA.

2 2 Theorem 1.5.4. Every distance preserving function f : R → R can be written uniquely as the composition of an orthogonal linear transformation followed by a translation. In particular, every such f is an isometry.

2 2 2 Proof. Set t := −f(0) ∈ R , and let ft : R → R be the corresponding translation. Then 2 2 the composition ftf : R → R is a distance preserving function satisfying f(0) = 0, so by the previous lemma we have ftf = g for some orthogonal linear transformation 2 2 g : R → R , and hence f = f−tg displays f as a composition of an orthogonal linear transformation followed by a translation. The uniqueness of this expression is established in the same manner one establishes uniqueness of the corresponding expression for aﬃne transformations in §1.4.

2 2 Corollary 1.5.5. Every distance preserving function f : R → R is an aﬃne transformation.

The following corollary could also be proved directly:

2 2 Corollary 1.5.6. Every isometry f : R → R can be expressed as a composition of 2 reﬂections (Example 1.3.6), hence the group Isom(R ) can be generated by elements of order 2.

Proof. In light of the theorem it suffices to show that every translation and every orthogonal transformation can be expressed as a composition of reflections. Given two 0 parallel lines L, L , the composition rL0 rL of the corresponding reflections will be the 2 0 translation f2t, where t ∈ R is the vector such that L = ft(L) (exercise!), so by choosing L and L0 appropriately, we can realize every translation as a composition of (at most) two reflections. 2 In Example 1.3.6 we noted that every orthogonal translation of R is either a rotation around the origin, or a reflection over the x-axis followed by such a rotation. It thus remains only to prove that the counter-clockwise rotation fθ around the origin through an angle θ can be expressed as a composition of reflections. Let L (resp. L0) be the line through the origin obtained by rotating the x-axis counter-clockwise around the origin through the angle θ/2 (resp. θ). 18 Chapter 1 Transformations of the Plane

I claim that the composition rL0 rL is fθ. As discussed in Example 1.3.6, one can see by linear algebraic methods that both rL and rL0 are orthogonal transformations of determinant −1, hence rL0 rL is an orthogonal transformation of determinant 1, so it must be a counter-clockwise rotation around the origin through some angle by the classiﬁcation of orthogonal transformations in Proposition 1.3.3. This reduces us to showing that (rL0 rL)(e1) is equal to fθ(e1), which is (cos θ, sin θ). This is exactly what 0 we arranged by our choice of L and L : We have rL(e1) = fθ(e1) by the choice of L, and 0 0 this point is ﬁxed by rL0 since it lies on L , by our choice of L .

2 2 Remark 1.5.7. The proof of the above corollary shows that every isometry f : R → R is the composition of at most ﬁve reﬂections. In fact at most three are needed.

2 In light of the classiﬁcation of isometries of R in Theorem 1.5.4, we can make the following deﬁnition, which will be useful later:

2 2 Definition 1.5.8. An isometry f : R → R is called orientation preserving iff, in the unique expression f = ftg of f as a composition of an orthogonal transformation g followed by a transflation ft, the orthogonal transformation g is a special orthogonal linear transformation (i.e. a rotation—see Example 1.3.5).

Remark 1.5.9. It is possible to define the notion of “orientation preserving” for any n n homeomorphism f : R → R in a manner restricting to the notion in Definition 1.5.8 for 2 isometries of R and satisfying the expected properties: The identity map is orientation preserving and a composition fg of homeomorphisms is orientation preserving iff either both f and g are orientation preserving, or neither f nor g is orientation preserving.

1.6 Algebraic automorphisms

n m Let K be a ﬁeld. Let us agree that a function f : K → K is algebraic iﬀ there are polynomials f1(x1, . . . , xn), . . . , fm(x1, . . . , xn) ∈ K[x1, . . . , xm] such that

f(x1, . . . , xn) = (f1(x1, . . . , xn), . . . , fm(x1, . . . , xn))

n n n for every x = (x1, . . . , xn) ∈ K . A function f : K → K is called an algebraic n −1 automorphism of K iff f is bijective and both f and f are algebraic. For example, every affine transformation is an algebraic automorphism. One can 1 ∗ check that every algebraic automorphism of K is of the form x 7→ λx+a for some λ ∈ K , 1 a ∈ K—in other words, every algebraic automorphism of K is an affine transformation. 2 2 2 For K , however, there are algebraic automorphisms K → K which are not affine transformations:

2 2 2 Example 1.6.1. The function f : K → K defined by f(x, y) := (x, x + y) is an 2 2 algebraic automorphism of K with inverse g given by g(x, y) = (x, x − y). The reader can check as an exercise that f is not an affine transformation (unless K = F2 is “the” 2 field with two elements, for then x = x for “every” (i.e. “both”) x ∈ F2). 1.7 Exercises 19

The Jacobian Conjecture is a famous unsolved conjecture about algebraic automor- n phisms of C . It says that if f1, . . . , fn ∈ C[x1, . . . , xn] are polynomials such that the polynomial   ∂f1 ··· ∂f1 ∂x1 ∂xn  . .  det  . .    ∂fn ··· ∂fn ∂x1 ∂xn (the Jacobian determinant) is a non-zero constant, then the corresponding algebraic n n function f : C → C is an algebraic automorphism. As mentioned above, this is not hard to see when n = 1, though it is unknown even when n = 2.

1.7 Exercises

2 2 Exercise 1.1. Let f : R → R be the linear transformation corresponding to the (invertible) matrix ! 1 1 A = . 0 1

1 2 1 Let S := {x ∈ R : d(x, 0) = 1} be the unit circle. Draw a picture of the image f(S ) of S1 under f and a picture of f(f(S1)).

2 2 Exercise 1.2. Let f : R → R be the linear transformation corresponding to the matrix ! cos θ − sin θ Aθ := sin θ cos θ

(for some θ ∈ R/2πZ). Show, by explicit calculation with the distance formula, that f is an isometry.

Exercise 1.3. Prove that the special orthogonal group SO2 is a normal subgroup of the orthogonal group O2.

2 2 Exercise 1.4. For points x, y ∈ R , the line segment xy is the set of points z in R that 2 can be written in the form z = tx + (1 − t)y for some t ∈ [0, 1]. Prove that a point z ∈ R lies on the line segment xy iﬀ the triangle inequality d(x, y) ≤ d(x, z) + d(z, y) is actually an equality.

2 2 2 Exercise 1.5. Suppose f : R → R is an isometry and let L ⊆ R be a line. Prove 2 2 “directly” (without making use of the classiﬁcation of isometries of R ) that f(L) ⊆ R is 2 a line. For clarity, let us take as the deﬁnition of a line: a subset L ⊆ R for which there exist real numbers a, b, c (with a, b not both equal to zero) such that

2 L = {x = (x1, x2) ∈ R : ax1 + bx2 + c = 0}. (The previous exercise might be helpful.) Chapter 2

Parametric Plane Curves

2 There are many possible things one can mean by a “curve” in the plane R . In this chapter we will return to the study of parametrized plane curves, which should be familiar from calculus classes. Here are the main deﬁnitions:

2 Definition 2.0.1. A parametrized plane curve is a function γ :[a, b] → R , defined on some closed interval [a, b] ⊆ R. We use t to denote a point of [a, b] and x, y :[a, b] → R to denote the coordinates of γ, so that γ(t) = (x(t), y(t)). We assume that the derivatives x0, y0, x00, y00 with respect to t exist and are continuous. We call γ0 := (x0, y0) the tangent vector to γ and we say that γ is smooth iff γ0(t) 6= (0, 0) for all t ∈ [a, b]. A smooth parametrized plane curve γ is said to be parameterized by arc length iff a = 0 and |γ0(t)| = 1 for all t ∈ [0, b]—in this case we generally use s instead of t for the parameter and we write ` instead of b, for reasons that will become clear momentarily.

Example 2.0.2. Fix a positive real number r. The function

2 γ : [0, 2π] → R γ(t) := (r cos t, r sin t) is a smooth parametrized curve whose image γ([0, 2π]) is a circle of radius r centered at the origin. The function

2 η : [0, 2πr] → R η(s) := (r cos(s/r), r sin(s/r)) is a smooth parametrized plane curve parametrized by arc length with the same image as γ.

In this chapter, we will use the word curve (resp. curve parametrized by arc length) to mean smooth parameterized plane curve (resp. smooth parametrized plane curve parametrized by arc length). We will study various geometric quantities associated with curves—for example, the length. Our assumption that the derivatives x0 and y0 exist and are continuous ensures that we can deﬁne the length in a fairly simple manner. While there are much more general deﬁnitions of length, they are less amenable to calculation

20 2.1 Reparametrization 21 and, in any case, are beyond the scope of these notes. The other important invariant attached to a curve is its curvature. Since our curves are plane curves, one can actually attach a sign to the curvature (this is not usually done in calculus classes) to deﬁne the signed curvature of a curve γ—this is a continuous function k :[a, b] → R. This is ensured by our assumption that the second derivatives x00 and y00 exist and are continuous. Here I want to focus on the interaction between the group of isometries of the plane and the properties of plane curves. The upshot is that the length and curvature are isometry invariant and they are, in some sense we will make precise, the “only” isometry invariants of a curve.

2.1 Reparametrization

The reader will probably agree that the two curves γ, η in Example 2.0.2 are in some sense “the same.” After all, the function γ is just the composition of multiplication by r, viewed as a function [0, 2π] → [0, 2πr], followed by η. In this section we are going to make this notion of “the same” precise. This point is implicit in the calculus treatment of parametrized curves, but it never really made explicit, probably because it is a bit technical. Consider two closed intervals [a, b], [c, d] ⊆ R. Let us use s for the coordinate on [c, d] and t for the coordinate on [a, b] for clarity. Let us agree that a nice bijection g :[c, d] → [a, b] is a bijection g :[c, d] → [a, b] which is twice continuously differentiable with g0(s) > 0 for all s ∈ [c, d]. It follows from the Implicit Function Theorem (see Theorem 6.3.1 in the appendix for a precise statement) that the inverse of g is also a nice bijection. Let us discuss this point a bit further, at it is rarely explained well in an introductory calculus sequence. For clarity, let us denote the inverse of g by f = f(t):[a, b] → [c, d], rather than by g−1. Then we have fg = Id, or, in other words, f(g(s)) = s for every s ∈ [c, d]. If we differentiate both sides of this equality with respect to s by using the Chain Rule, we find that

f 0(g(s))g0(s) = 1. (2.1)

Then, using the fact that g0(s) > 0, we solve (2.1) for f 0(g(s)) to find f 0(g(s)) = 1/g0(s). Since every t ∈ [a, b] is g(s) for some (in fact a unique) s ∈ [c, d], we conclude that f 0(t) > 0 for all t ∈ [a, b]. Notice that we have to use the Implicit Function Theorem here to know a priori that f is differentiable, so that the Chain Rule can be applied to calculate the derivative of the composition f(g(s)). We can use the same method to calculate the second derivative of g, since we know from the Implicit Function Theorem that it exists (and is continuous): We just differentiate (2.1) again with respect to s to find

f 0(g(s))g00(s) + g0(s)2f 00(g(s)) = 0 and then we use that g0(s) > 0 to solve for f 00(g(s)). This gives f 0(g(s))g00(s) f 00(g(s)) = − . g0(s)2 22 Chapter 2 Parametric Plane Curves

The Implicit Function Theorem says that this formula is indeed valid, so f 00 exists and is continuous because g00 exists and is continuous. It is customary in the situation described above to get rid of the notation for g and its inverse, and just view t = t(s) as a function of s, and s = s(t) as a function of t. With this understanding, formula (2.1) is rewritten ds dt = 1. (2.2) dt ds 2 Definition 2.1.1. If γ :[a, b] → R is a (smooth parametrized plane) curve and 2 t(s):[c, d] → [a, b] is a nice bijection, then the composition γt :[c, d] → R is also a curve (by the Chain Rule). A curve obtained from γ in this manner is called a 2 2 reparametrization of γ. Two curves γ :[a, b] → R , η :[c, d] → R are called equivalent iff η is a reparametrization of γ.A parametric plane curve is an equivalence class of smooth parametrized plane curves. The notion of “equivalence” defined above is indeed an equivalence relation on the set of curves. The observation at the beginning of this section implies that the two curves in Example 2.0.2 are equivalent.

2.2 Length and the arc length parametrization

Among all reparametrizations of a given curve, there is one particularly nice choice, called the arc length parametrization, which we will now describe. 2 0 0 0 0 Given a curve γ = (x, y):[a, b] → R , recall that we call γ = γ (t) = (x (t), y (t)) the tangent vector to γ at γ(t). The smoothness assumption on γ says that γ0 is never the zero vector, hence its magnitude p |γ0| = (x0)2 + (y0)2

0 is a function |γ | :[a, b] → R>0. We deﬁne the length ` = `(γ) of γ to be Z b `(γ) := |γ0(t)|dt. (2.3) a More generally, we deﬁne the arc length function s :[a, b] → [0, `] by Z t s(t) := |γ0(τ)|dτ. a By the Fundamental Theorem of Calculus we have ds = |γ0(t)|, (2.4) dt which is > 0 by the assumption that γ is smooth. Our assumption that x00 and y00 exist and are continuous also shows that d2s x0x00 + y0y00 = dt2 |γ0(t)| 2.3 Curvature and the Frenet formulas 23 exists and is continuous, hence our arc length function s is a nice bijection, and hence so is its inverse

t = t(s) : [0, `] → [a, b].

As discussed in §2.1, we have

dt 1 = (2.5) ds ds/dt 1 = . |γ0(s(t))|

2 The reparametrization η(s) := γ(t(s)) : [0, `] → R of γ is called the arc length parametrization of γ. As the terminology suggests, the curve η is parameterized by arc length because we compute

dη dγ dt = ds dt ds

dγ dt = dt ds = 1 using the Chain Rule and (2.5). The upshot of this section is that if one considers curves up to reparametrization, then nothing is lost (in theory) by considering only curves parametrized by arc length. However, it should be mentioned that the integrals deﬁning `(γ) and s(t) can rarely be carried out in terms of elementary functions, so if one ever needs to explicitly compute a quantity associated to the arc length parametrization of a curve γ, one usually tries to express that quantity in terms of the original parametrization of γ by using the Chain Rule.

2.3 Curvature and the Frenet formulas

2 0 Consider a curve γ = γ(s) : [0, `] → R parametrized by arc length. We call T(s) := γ (s) the unit tangent vector to γ at γ(s). Since γ is parametrized by arc length, we have |T| = 1, independent of s. By linear algebra, there is a unique unit vector N = N(s) orthogonal to T such that the matrix (T, N) with columns given by T and N has determinant 1. The vector N(s) is called the principal unit normal to γ at γ(s). Since |T| = 1 is constant, if 0 0 we diﬀerentiate both sides with respect to s we ﬁnd that T · T = 0. So T is orthogonal to T, hence it must be some scalar multiple of N:

0 T = kN. (2.6)

The scalar multiple k = k(s) is called the signed curvature of γ at γ(s). The formula for k in Exercise 2.3 makes it clear that k is a continuous function k : [0, `] → R. 24 Chapter 2 Parametric Plane Curves

0 One can now consider the derivative N . By the same arguments used for T, we 0 0 see that N will be some scalar multiple of T: N = αT. In fact this α doesn’t give us anything “new”: By diﬀerentiating hN, Ti = 0, we ﬁnd that

0 0 hN , Ti + hN, T i = 0.

Using the deﬁnitions of the signed curvature k and α we can rewrite this

hαT, Ti + hN, kNi = 0.

Using bilinearity of the inner product and the fact that T and N are unit vectors, this gives α + k = 0, or α = −k. The discussion thus far establishes the Frenet formulas (for curves in the plane parametrized by arc length):

0 T = kN (2.7) 0 N = −kT.

2 Example 2.3.1. Consider the curve parametrized by arc length η : [0, 2πr] → R from 2 Example 2.0.2 whose image is the circle of radius r centered at the origin in R . Note that η traces out its image in the counter-clockwise direction. We have

0 T = (r cos(s/r), r sin(s/r)) = (− sin(s/r), cos(s/r)).

Since ! − sin(s/r) − cos(s/r) cos(s/r) − sin(s/r) is an orthogonal matrix, we have

N = (− cos(s/r), − sin(s/r)).

We compute

0 T = (−(1/r) cos(s/r), −(1/r) sin(s/r)) = (1/r)N, so the signed curvature k of η is constant, equal to 1/r.

If there is any chance of confusion (for example, if we consider two curves parametrized by arc length at the same time) then we will write Tγ, Nγ, and kγ instead of T, N, and k. As mentioned in the previous section, the arc length parametrization of a curve γ = γ(t) is generally difficult, or—in some sense—impossible, to write down, so it is worth having formulas for T, N, and k in terms of an arbitrary parametrization. This is done by using formula (2.5), together with the Chain Rule, to re-express the derivatives 2.4 Plane curves up to isometry 25 with respect to arc length s used to define T, N, and k in terms of differentiation with respect to t. For example,

dγ(t(s)) (t) = T ds dγ dt = dt ds 1 = γ0(t), |γ0(t)| where the primes denote differentiation with respect to t. Then N(t) is defined from T(t) in the same way that N(s) is defined from T(s). Similarly we compute d d dt T = T ds dt ds 1 = 0, |γ0|T where the primes denote derivatives with respect to t. Then k(t) is defined by

d T = k(t) (t). ds N

2.4 Plane curves up to isometry

2 2 If f : R → R is any nice enough function (a smooth bijection with smooth inverse, say), 2 2 and γ :[a, b] → R is a smooth parametrized curve, then the composition fγ :[a, b] → R is also a smooth parametrized curve. We will be interested in this construction only 2 when f is an orientation preserving isometry (Deﬁnition 1.5.8) and γ : [0, `] → R is a curve parametrized by arc length. In this situation we can write f = ftg, where g ∈ SO2 and ft is a translation. Let us write A = Ag for the orthonormal matrix corresponding to g. One checks (Exercise 2.4) that:

0 0 1. (fγ) = Aγ , hence fγ is also parameterized by arc length and Tfγ = ATγ.

2. `(fγ) = `(γ).

3. The principal unit normal vectors of γ and fγ are related by Nfγ = ANγ.

4. The curvatures of γ and fγ are equal: kγ = kfγ.

2 2 Theorem 2.4.1. Suppose γ : [0, `] → R and γ : [0, `] → R are two curves parametrized by arc length. Then the following are equivalent:

2 2 1. γ = fγ for some orientation preserving isometry f : R → R .

2. The curves γ and γ have the same length and curvature: That is ` = ` and k = k. 26 Chapter 2 Parametric Plane Curves

Proof. The fact that (1) implies (2) is discussed above and left for the reader as Exer- cise 2.4. To see that (2) implies (1), suppose γ and γ are curves parametrized by arc 2 length with the same length and curvature. Let t := γ(0) − γ(0) ∈ R and let A ∈ SO2 0 0 be the rotation taking the unit vector γ (0) to the unit vector γ (0). Then f := ftfA is 2 an isometry of R such that the curves γ and fγ have the same initial point and initial tangent vector. By the implication already “proved,” these two curves also have the same length and curvature. By replacing γ with fγ, we thus reduce to proving that if γ and γ are curves parametrized by arc length with the same initial point, the same initial tangent vector, and the same length ` and curvature k, then they are equal. Let T and N (resp. T and N) be, respectively, the unit tangent vector and principal unit normal for the curve γ (resp. γ). Using the Frenet Formulas (2.7) we compute 0 hT − T, T − Ti + hN − N, N − Ni 0 0 0 0 = 2hT − T , T − Ti + 2hN − N , N − Ni = 2hkN − kN, T − Ti − 2hkT − kT, N − Ni = −2khN, Ti − 2khN, Ti + 2khN, Ti + 2khN, Ti = 0. This shows that the function in parentheses is constant—but it is zero at s = 0 since T(0) = T(0) (hence N(0) = N(0) as well), so it is zero. But this function is the sum of the magnitude of T − T and the magnitude of N − N, both of which are non-negative, so 0 0 we must have T = T. Since T = γ and T = γ , this proves that the derivatives of the coordinate functions of γ and γ are equal. But we also have γ(0) = γ(0), hence γ = γ by the Fundamental Theorem of Calculus. 3 Remark 2.4.2. There is an analogue of Theorem 2.4.1 for curves in R parametrized by arc length. In this case, the curvature cannot be given a sign, and one must also consider another invariant called the torsion. (The proof is essentially the same, though the Frenet formulas become a bit more complicated in three dimensions.) For this, see the ﬁrst chapter of the book Diﬀerential Geometry of Curves and Surfaces by Manfredo P. Do Carmo, from which the material in this section has been “borrowed.”

2.5 Exercises

2 2 2 Exercise 2.1. Suppose f : R → R is an isometry of R which is orientation reversing 2 (not orientation preserving) and γ is a curve parametrized by arc length in R . Show 2 that fγ is a curve parametrized by arc length in R with signed curvature kfγ = −kγ. 2 Exercise 2.2. Let P and Q be distinct points of R . Let v := (Q − P )/|Q − P | be the unit vector pointing from P to Q and let ` := |Q − P | be the distance from P to Q. Show 2 that the curve γ : [0, `] → R deﬁned by γ(s) := P + sv is a smooth curve parametrized by arc length whose signed curvature k is identically zero. 2 Exercise 2.3. Consider a smooth plane curve γ = γ(s) = (x(s), y(s)) : [0, `] → R parametrized by arc length. Show that the signed curvature k of γ is given by k = x0y00 − x00y0. 2.5 Exercises 27

Hint: First figure out an explicit formula for the principal unit normal N in terms of x, y 0 and their derivatives. You also have an obvious such formula for T , so you just have to 0 check that the proposed formula for k actually satisfies T = kN. For this, it might be helpful to have the identity obtained by writing out the condition |T| = 1 in terms of x, y and their derivatives then differentiating both sides.

Exercise 2.4. Check the four listed assertions at the beginning of §2.4. Chapter 3

Aﬃne Algebraic Plane Curves

2 Generally speaking, a curve should be a subset of the plane R obtained by imposing 2 some “reasonable” relationship between the two coordinates x, y of a point (x, y) ∈ R . 2 For example, the graph Γf ⊆ R of a function f : R → R is a curve since it is obtained by requiring that the second coordinate be the value of f on the ﬁrst coordinate:

2 Γf := {(x, y) ∈ R : y = f(x)}. It seems reasonable then, to say that an algebraic curve (more precisely, an aﬃne algebraic 2 plane curve) is a subset of R obtained by imposing an “algebraic” (i.e. polynomial) relationship between the coordinates. For example, if

n f(x) = a0 + a1x + ··· + anx

2 is a polynomial function of x, then the graph Γf ⊆ R is an algebraic curve. Here are some other algebraic curves that you should have encountered at some point:

2 2 2 C1 := {(x, y) ∈ R : x + y = 1} 2 C2 := {(x, y) ∈ R : x = 0} 2 2 2 C3 := {(x, y) ∈ R : x − y = 1}

The curve C1 is a circle of radius one centered at the origin, C2 is the y-axis, and C3 is a hyperbola. Of course, not all plane curves are graphs of functions (e.g. C1, C2, C3 above). The notion of an algebraic curve makes sense over any field K. For example, replacing R everywhere with the complex numbers C, one can also consider complex algebraic 2 2 curves C ⊆ C in the complex plane C . It may seem at first that this might complicate matters: For one thing, it becomes difficult to visualize a complex plane curve, since 2 the complex plane C itself has “four real dimensions.”On the other hand, the field of complex numbers C has certain technical advantages over R, particularly the fact that it is algebraically closed: Any polynomial (of positive degree in one variabel) with complex coefficients can be factored as a product of linear (degree one) polynomials. In general, the theory of algebraic curves, as we have defined them, is most satisfactory over an algebraically closed field, where the relationship between a polynomial and the algebraic curve it defines can be most satisfactorily understood.

28 3.1 Polynomials and zero loci 29

As this course is meant to be a course on projective geometry, we will not dwell overly on the subject of affine curves. Our purpose in this section is mainly just to give the reader a taste for the theory of algebraic plane curves, without with our later study of projective plane curves would seem rather unmotivated. In order for this chapter to have some actual content, I have chosen to discuss the problem of classifying affine conics (over R and C) up to an appropriate notion of affine equivalence. The classification will be given without proof, though we shall give a complete treatment of the analogous classification problem in the projective setting. Already in this classification problem we will see that C has an important “advantage” over R: Every complex number is a square, whereas real numbers less than zero are not squares.

3.1 Polynomials and zero loci

Recall that a polynomial in a single variable x with coefficients in a field K is a formal expression f(x) of the form 2 n f(x) = a0 + a1x + a2x + ··· + anx , where the coefficients ai are elements of K. The set of polynomials with coefficients in K is denoted K[x]. This set carries various additional structure: It is an K-algebra, which basically means that one can add, subtract, and multiply polynomials in a meaningful way, and that one can regard an element of K as a constant polynomial. Though it is perhaps a stretch to emphasize this point now, it might be worth explaining—for the more mathematically sophisticated student—what is meant by “formal expression.” We really mean here that a polynomial is just a way of writing (and thinking about) a sequence (a0, a1, a2,... ) of elements of K which is eventually zero. We can add, subtract, and multiply such sequences (by thinking of them as coefficients of polynomials). Now, of course, we can also view the polynomial as a function f : K → K, as we often do in calculus when K = R. It might not seem like such a big deal to distinguish between the formal list of coefficients and the corresponding function, and, indead, it is not in many senses: you can recover the polynomial (i.e. its list of coefficients) from the corresponding function f : R → R by the rule f (n)(0) a = n n! (as you learn in the study of Taylor series). On the other hand, if you work with coefficients in a finite field K, say, then of course we can still view a polynomial f(x) ∈ K[x] as a function f : K → K, but obviously there can be no hope of recovering the polynomial from the corresponding function because K[x] is infinite, while there are only finitely many functions K → K)! For example, if K = F2 = Z/2Z = {0, 1} is the two-element field, then 2 2 2 x = x as functions K → K, but we do not think of x = 0 + 1x and x = 0 + 0x + 1x as being the same polynomial. This distinction is largely a pedagogical matter, but it might be wise for the reader to forget everything he/she knows about functions R → R for a while and just think about polynomials in a completely formal way. 30 Chapter 3 Affine Algebraic Plane Curves

Deﬁnition 3.1.1. The degree of a polynomial f as above is the largest n for which the coeﬃcient of xn in f is non-zero:

deg(f) := max{i : ai 6= 0}.

Thus a non-zero constant polynomial f(x) = a0 6= 0 has degree zero. It is often convenient to deﬁne the degree of the zero polynomial to be −∞. The terms “non-constant” and “of positive degree” are equivalent; similarly “linear” and “degree one” are interchangeable.

We can make the same deﬁnitions for polynomials in two (or more) variables x, y. Now a polynomial is a formal expression of the form

∞ X i j f(x, y) = ai,jx y , i,j=0 where all but ﬁnitely many of the real numbers aij are zero. The degree of such a polynomial is deﬁned by

deg(f) := max{i + j : aij 6= 0}.

Evidently, any polynomial f(x, y) of degree d can be written

X i j f = ai,jx y i+j≤d with at least one of ad,0, ad−1,1, . . . , a1,d−1, a0,d non-zero. The set K[x, y] of polynomials in variables x, y forms an K-algebra.

Deﬁnition 3.1.2. Given a polynomial f(x, y) ∈ K[x, y], we call the subset

2 Z(f) := {(x, y) ∈ K : f(x, y) = 0}

2 2 of K the zero locus of f(x, y). A subset of K of the form Z(f), for some non-constant f(x, y) ∈ R[x, y] will be called an aﬃne algebraic plane curve over K, or simply, an 2 algebraic curve. If C ⊆ K is an algebraic curve, then any f ∈ K[x, y] for which C = Z(f) will be called a deﬁning equation for C.

2 When discussing algebraic curves, we often call K the aﬃne plane over K. Usually 2 2 we refer to R simply as the plane and to C as the complex plane.

Remark 3.1.3. If the field K is contained in a larger field K, then of course we can also consider the zero locus of a polynomial f(x, y) ∈ K[x, y] in the larger affine plane 2 K . If there is any chance of confusion, we will add some notation to Z(f) to emphasize 2 which field we are working with. We will write Z(f)(K) for the zero locus of f in K , to 2 distinguish it from the zero locus of f in K , which we will denote Z(f)(K). The interested student may note that the Galois group G = Gal(K/K) (the group of automorphisms of the field K which restrict to the identity on the subfield K ⊆ K) acts on Z(f)(K). The 3.1 Polynomials and zero loci 31

student who wants to think about this at all should think about the case where the field extension is R ⊆ C, in which case the Galois group is the group with two elements because complex conjugation z 7→ z is the unique non-identity automorphism of C restricting to 2 2 the identity on R. If f ∈ R[x, y] and (a, b) ∈ Z(f)(C) ⊆ C , then (a, b) ∈ Z(f)(C) ⊆ C as well. This is because, for any a, b ∈ C, we have f(a, b) = f(a, b) since the coefficients of f are real. It was realized early on in the development of algebraic geometry that the best “geometric” picture of the polynomial f is obtained by considering all the sets Z(f)(K), for all extension fields K ⊆ K, together with the actions of the relevant Galois groups. In modern algebraic geometry this information is all packaged together by considering the ring K[x, y]/(f) and its associated “prime spectrum.” The historical development of this perspective is described in [D].

2 For example, the subsets C1, C2, C3 of R from the introduction to this chapter can be 2 2 2 2 written Z(f1), Z(f2), Z(f3) for f1(x, y) = x + y − 1, f2(x, y) = x, f3(x, y) = x − y − 1. Since a product of two elements of K is zero iﬀ at least one of them is zero, we have

Z(fg) = Z(f) ∪ Z(g).

For example, Z(xy) is the union of the y-axis and the x-axis. The subset Z(f) does not determine the polynomial f: It is certainly possible that ∗ Z(f) = Z(g) for diﬀerent polynomials f, g. For example, if λ ∈ K = K \{0}, then clearly Z(f) = Z(λf) since f(x, y) = 0 iﬀ λf(x, y) = 0. For similar reasons,

Z(f) = Z(f 2) = Z(f 3) = ··· .

2 2 “Even worse,” there are lots of polynomials f ∈ R[x, y], such as f(x, y) = x + y + 1, 2 2 2 for which Z(f) ⊆ R is empty, and lots of polynomials such as f(x, y) = x + y , or f(x, y) = x4 + y4, for which Z(f) consists of a single point. At this point one might wonder whether “algebraic curve” is really a good name for the sets Z(f) which can be finite (or empty), despite being called “curves,” which suggests they are one dimensional in some sense. One such sense is that the Krull dimension of the ring K[x, y]/(f) is 1 for every non-constant f(x, y) ∈ K[x, y] (by Krull’s Hauptidealsatz). Furthermore, the examples where Z(f) is empty or a finite set of points occur only when working over the real numbers R. Over C, it turns out that, after possibly removing finitely many points from Z(f), the set Z(f) carries, in some reasonably natural way, the structure of a (non-empty!) complex one dimensional manifold. We will not use these facts in anything that follows; they are mentioned only to justify our terminology. Over the complex numbers, one can, roughly speaking, recover f from the subset 2 Z(f) ⊆ C up to the sort of ambiguities discussed two paragraphs above. The precise statement is Hilbert’s Nullstellensatz which says that the subset (ideal, in fact)

2 {g(x, y) ∈ C[x, y]: g(x, y) = 0 for all (x, y) ∈ Z(f) ⊆ C }

of C[x, y] coincides with the subset (ideal)

n {g(x, y): g = hf for some h ∈ C[x, y] and some n ∈ {1, 2,... }}. 32 Chapter 3 Aﬃne Algebraic Plane Curves

We mention this only for the sake of completeness; we will not use this fact elsewhere. Since the subset Z(f) is determined from f, but not vice-versa, we often think of the polynomial f as being the more fundamental object. This point of view will be especially important in §3.3 where we discuss the problem of “classifying” the algebraic plane curves (of, say, a given degree). 2 In particular, we would like to say that the degree of an algebraic plane curve C ⊆ K is the degree of “the” polynomial f for which C = Z(f). Of course this makes no sense because there are many such polynomials f! There are various ways of resolving this—we could, for example, consider the minimum degree of all those f for which C = Z(f). Rather than worry about this technicality, let us just agree to the following:

2 Deﬁnition 3.1.4. A line (resp. conic, cubic, quartic, . . . ) is an algebraic curve C ⊆ K which can be written C = Z(f) for some polynomial f ∈ K[x, y] of degree one (resp. two, three, four, . . . ).

Notice that, according to this deﬁnition, every “line” is also a conic, because if C = Z(f) with f linear, then we also have C = Z(f 2), and f 2 will have degree two.

3.2 Smooth and singular points

Just as there is a notion of “smoothness” for parametric curves (Chapter 2), there is a very closely related notion of smoothness for algebraic curves. Notice that the partial derivatives fx and fy of a polynomial f ∈ K[x, y] make sense over any ﬁeld K (not just over the real numbers, where they have a “calculus” interpretation in terms of limits). Explicitly, if

X i j f = ai,jx y i,j then we deﬁne fx and fy to be

X i−1 j fx := iai,jx y i,j X i j−1 fy := jai,jx y . i,j

It is perhaps not so obvious that these “definitions” of the derivative should have any geometric significance over other fields, though we shall see that this is the case.

2 Definition 3.2.1. For f ∈ K[x, y], a point P = (x0, y0) ∈ Z(f) ⊆ K is called a non- singular (or smooth) point of Z(f) iff at least one of fx(P ), fy(P ) is non-zero. A point P ∈ Z(f) for which fx(P ) = fy(P ) = 0 is called a singular point of Z(f). The subset of Z(f) consisting of smooth (resp. singular) points is denoted Z(f)sm (resp. Z(f)sing). We sing say that f is non-singular or smooth iff Z(f) (K) is empty for every field extension 1 K ⊆ K.

1 sing It equivalent to ask that Z(f) (K) = ∅ for some algebraically closed ﬁeld K containing K. 3.3 Change of variables and aﬃne equivalence 33

It is worth emphasizing that the question of whether P ∈ Z(f) is a smooth point or a singular point depends highly on f, not just on the curve C = Z(f). For example, 2 2 for any (non-constant) f ∈ K[x, y], we have Z(f) = Z(f ), but every point of Z(f ) is a singular point (Exercise 3.2). sing In defining smoothness, we require that Z(f) (K) be empty for all field extensions K ⊇ K (rather than for just K itself) for various reasons. For one thing, if we have a polynomial f ∈ K[x, y] and we regard it as a polynomial in f ∈ K[x, y], we don’t want the meaning of “smooth” to change! A good example to keep in mind is the polynomial 2 2 f = (1 + x ) ∈ R[x, y], which has empty zero locus over R, even though every point of Z(f)(C) is singular (c.f. Exercise 3.2). This phenomenon illustrates the general philosophy of Remark 3.1.3 that one should consider all zero loci of f over all field extensions of K together.

Deﬁnition 3.2.2. If P = (x0, y0) is a non-singular point of Z(f), we deﬁne the tangent line to Z(f) at P to be the line

TP Z(f) := Z(xfx(P ) + yfy(P ) − x0fx(P ) − y0fy(P )).

Notice that the assumption that P is non-singular is needed to ensure that the linear polynomial whose zero locus deﬁnes TP Z(f) is non-constant, so that TP Z(f) is actually a line. Note also that the constant term of the linear polynomial is chosen so that P ∈ TP Z(f).

2 Proposition 3.2.3. Let P = (a, b) be a smooth point of a curve Z(f) ⊆ R . Then there are open subsets U, V of R, containing a and b, respectively, such that Z(f) ∩ (U × V ) is either the graph of a smooth (inﬁnitely diﬀerentiable) function g : U → V (this will be the case if fx(P ) 6= 0) or the graph of a smooth function h : V → U (if fy(P ) 6= 0). In other words, at least one of the projections

π1 : Z(f) ∩ (U × V ) → U

π2 : Z(f) ∩ (U × V ) → V will be a continuous bijection with smooth inverse.

Proof. This is a special case of the Implicit Function Theorem (Theorem 6.3.1 in the appendix).

3.3 Change of variables and aﬃne equivalence

2 Recall the group Aff(K ) of affine transformations of the plane from §1.4. Given [A, t] ∈ 2 Aff(K ) and f ∈ K[x, y], we define a new polynomial f · [A, t] ∈ K[x, y] by setting

(f · [A, t])(x, y) := f([A, t](x, y)).

To be careful here, we should stick more precisely to the conventions of linear algebra 2 and write every element of K , as well as the argument of f, as a column vector. Then 34 Chapter 3 Aﬃne Algebraic Plane Curves we would write: ! ! x x (f · [A, t]) := f(A + t). y y To be completely explicit, if ! ! a b t A = and t = 1 , c d t2 then

f(x, y) = f(ax + by + t1, cx + dy + t2). Usually I will be more sloppy with the notation since it is painful to write column vectors in the inline text. A good compromise is to write x (or x if you prefer) for the vector of variables (x, y) (thought of as a column vector, but often abusively written as a row vector). Then f · [A, t] is defined by (f · [A, t])(x) := f(Ax + t), (3.1) where the argument of f is thought of as a column vector. The above construction of f · [A, t] defines an action (more precisely, a right action) 2 of Aff(K ) on K[x, y]. This means that 2 1. f · Id = f, where Id = [Id, 0] is the identity element of Aff(K ), and 2. (f · [A, t]) · [A0, t0] = f · ([A, t][A0, t0])

0 0 2 for every f ∈ K[x, y], [A, t], [A , t ] ∈ Aff(K ). The first equality is obvious and the second is a simple calculation using definition 2 (3.1) and the formula (1.3) for products in Aff(K ): ((f · [A, t]) · [A0, t0])(x) = (f · [A, t])(A0x + t0) = f(A(A0x + t0) + t) = f(AA0x + At0 + t) = (f · [AA0, At0 + t])(x) = (f · ([A, t][A0, t0]))(x).

2 One can also think of this action by noting that any [A, t] ∈ Aﬀ(K ) gives rise to an automorphism of the K-algebra K[x, y] given by

x 7→ ax + by + t1

y 7→ cx + dy + t2, where we have written out A and t as above. n More, general, the same formula (3.1) may be viewed as defining an action of Aff(K ) on K[x1, . . . , xn] (think of x as the column vector of variables (x1, . . . , xn)). 2 2 The actions of Aff(K ) on K and K[x, y] are “compatible” in various ways. For now, let us just note: 3.3 Change of variables and affine equivalence 35

Lemma 3.3.1. For f ∈ K[x, y], we have

Z(f · [A, t]) = [A, t]−1 · Z(f) = {[A−1, −A−1t](x): x ∈ Z(f)}.

2 Proof. A point x ∈ K is in Z(f ·[A, t]) iff f(Ax+t) = 0, which is true iff Ax+t = [A, t](x) is in Z(f), which is true iff

x = [A, t]−1 · ([A, t]x) ∈ [A, t]−1 Z(f).

In other words, the zero locus of f · [A, t] is obtained from the zero locus of f by 2 applying the inverse of the aﬃne transformation [A, t] to K . The reader should be familiar with the appearance of the “inverse” here, as this also occurs (for the same reason) when one thinks about, say, the relationship between the group of f(x) and the ∗ graph of f(ax + b), with a ∈ R , b ∈ R. Proposition 3.3.2. The degree of a polynomial is invariant under the action of aﬃne transformations. That is,

deg f = deg(f · [A, t])

2 for every f ∈ K[x, y] and every [A, t] ∈ Aﬀ(K ).

P i j Proof. Consider a degree d polynomial f(x, y) = i+j≤d aijx y . Then if we write out A and t as above, we have

X i j f · [A, t] = aij(ax + by + t1) (cx + dy + t2) . i+j≤d

From the rule for Multinomial Expansion and the fact that the degree of a product is the sum of the degrees of the factors, it is clear that (when i + j ≤ d) the polynomial i j (ax + by + t1) (cx + dy + t2) clearly has degree at most d. Since f · [A, t] is a sum of such things, we see that the degree of f · [A, s] cannot be any bigger than d = deg f:

deg(f · [A, t]) ≤ deg f.

But now we apply this same observation with the polynomial f replaced by the polynomial f · [A, t], and [A, t] replaced by its inverse to ﬁnd the opposite inequality (here we use the fact that we have an action to know that f = (f · [A, t]) · [A, t]−1).

Definition 3.3.3. Two polynomials f, g ∈ K[x, y] are called affine equivalent iff there 2 ∗ is an affine transformation [A, t] ∈ Aff(K ) and a non-zero scalar λ ∈ K such that λg = f · [A, t]. When this is the case, one says that g is obtained from f by affine linear 2 change of variables (and rescaling). Similarly, we say that two subsets S, T ⊆ K are 2 affine equivalent iff S = [A, t](T ) for some affine transformation [A, t] ∈ K . 36 Chapter 3 Affine Algebraic Plane Curves

2 One can check, using the fact that our construction of f · [A, t] is an action of Aff(K ) on K[x, y], that “being affine equivalent” is indeed an equivalence relation (Exercise 3.4). Notice that Lemma 3.3.1 (plus the fact that rescaling a polynomial doesn’t change its zero locus) implies that if f, g ∈ K[x, y] are affine equivalent, then their zero loci Z(f), 2 Z(g) are affine equivalent as subsets of K . The converse is certainly not true: we will soon see that there are inequivalent degree two polynomials f, g ∈ R[x, y] such that both Z(f) and Z(g) are empty! A very ambitious goal would now be to classify all polynomials and all plane curves up to affine equivalence. (That is, to “explicitly” describe all affine equivalence classes.) To attempt this, one has to think of many “affine invariants,” which is to say, “quantities 2 associated to a polynomial f in K[x, y] (or a subset C ⊆ K ) that depend only on the affine equivalence class of f (or C).” Proposition 3.3.2 gives us one such quantity, namely the degree of a polynomial. This lets us break up our classification problem degree by degree. As a warmup, let us classify linear (that is, degree one) polynomials up to affine equivalence. The corresponding classification of plane curves will be the classification of lines, in the sense of Definition 3.1.4. We will again take up the subject of lines, in the projective setting, in §5.4. We claim that in fact every degree one polynomial is equivalent to the polynomial x. A degree one polynomial is a polynomial of the form f(x, y) = Ax + By + C for some real numbers A, B, C with (A, B) 6= (0, 0). After applying the (invertible linear) transformation (x, y) 7→ (y, x) if necessary, we see that f is equivalent to a polynomial g of the same form where in fact A 6= 0. By applying the invertible linear transformation (x, y) 7→ (A−1x, y), we see that g is equivalent to a polynomial h of the form x + By + C. After applying the invertible linear transformation (x, y) 7→ (x − By, y), we see that h is equivalent to a polynomial j of the form x + C. Finally, applying the translation (x, y) 7→ (x − C, y), we see that j is equivalent to x. We leave it as an easy exercise for the reader to see that “a linear polynomial f(x, y) can be recovered (up to multiplication by a non-zero real number) from the corresponding plane curve (line) Z(f)”. We should also emphasize at this point that our notion of affine equivalence is not the only “reasonable” one: There are very legitimate reasons for rejecting the idea that plane curves differing by affine transformation are “equivalent.” For one thing, 2 affine transformations of R do not always distances, or even angles. It would be quite 2 reasonable to try to classify conics in R up to isometry (§1.2). While this can be done in an elementary manner, it is rather tedious and we shall not pursue this here. We will see this issue (the difference between affine equivalence and isometry-equivalence) manifested in our classification of conics in various ways: all ellipses are equivalent to the unit circle, all pairs of intersecting lines are equivalent (regardless of the angle of intersection), and so forth.

3.4 Five points determine a conic

2 If P and Q are distinct points of K , then it is easy to see that there is a unique line 2 L ⊆ K for which P,Q ∈ L. Here we prove an analogous statement for conics. The 3.4 Five points determine a conic 37 technique used in the proof will be recycled many times later.

2 Proposition 3.4.1. Suppose P1,...,P5 ∈ K are any ﬁve points, no three of which are collinear. Then, up to multiplying by a non-zero real number, there is a unique degree two polynomial f(x, y) ∈ K[x, y] such that P1,...,P5 ∈ Z(f).

Proof. Let us ﬁrst treat the special case where P1 = (0, 0), P2 = (1, 0), and P3 = (0, 1). The general degree two polynomial takes the form

f(x, y) = Ax2 + Bxy + Cy2 + Dx + Ey + F, where (A, B, C) 6= (0, 0, 0). The condition that (0, 0) ∈ Z(f) is equivalent to F = 0, and the conditions that (1, 0) ∈ Z(f) and (0, 1) ∈ Z(f) are then equivalent to D = −A and E = −C, so our f must take the form

f(x, y) = Ax2 + Bxy + Cy2 − Ax − Cy.

If we write P4 = (x1, y1), P5 = (x2, y2), then the conditions that P4,P5 ∈ Z(f) are equivalent to the system of linear equations

Ax1(x1 − 1) + Bx1y1 + Cy1(y1 − 1) = 0

Ax2(x2 − 1) + Bx2y2 + Cy2(y2 − 1) = 0.

By basic linear algebra, we can conclude that this system has a unique non-zero solution (A, B, C), up to scaling, provided that we can show the coeﬃcient vectors

(x1(x1 − 1), x1y1, y1(y1 − 1)) and (x2(x2 − 2), x2y2, y2(y2 − 1)) are linearly independent. Suppose not. Then there is some λ ∈ K so that

λ(x1(x1 − 1), x1y1, y1(y1 − 1)) = (x2(x2 − 2), x2y2, y2(y2 − 1)).

Now, since P4 isn’t on the line joining (0, 0) and (1, 0) (the x-axis), we have y1 6= 0. Since P4 isn’t collinear with (0, 0) and (1, 0), we have x1 6= 0. For similar reasons, we have x2, y2 =6 0. We can then solve for λ using equality of the second coordinates above to ﬁnd λ = (x2y2)/(x1y1). Plugging in for λ and cancelling the non-zero real number x2, equality of the ﬁrst coordinates above then yields:

y2(x1 − 1) = y1(x2 − 1).

But y(x1 − 1) = y1(x − 1) is the equation defining the line containing P4 = (x1, y1) and P2 = (1, 0), so this latter equality shows that P2,P4, and P5 are collinear—a contradiction. We deduce the general case from the special case as follows: By Proposition 1.4.3 we 2 2 can find an affine transformation [A, t]: K → K taking P1, P2, and P3 to (0, 0), (1, 0), and (0, 1), respectively. Since an affine transformation takes lines to lines, the five points

(0, 0) = [A, t](P1), (1, 0) = [A, t](P2), (0, 1) = [A, t](P3), [A, t](P4), [A, t](P5) 38 Chapter 3 Affine Algebraic Plane Curves are in general position (no three collinear), so by the special case treated above, there is a unique (up to rescaling) degree two f ∈ K[x, y] such that Z(f) contains these five points. By Proposition 3.3.2 and Lemma 3.3.1, the polynomial f · [A, t] is of degree two and its zero locus contains P1,...,P5. If g is any other degree two polynomial whose −1 zero locus contains P1,...,P5, then that proposition and lemma imply that g · [A, t] is a degree two polynomial whose zero locus contains the five points listed above, hence g · [A, t]−1 must be a rescaling of f by the special case treated above, and hence g must be a rescaling of f · [A, t].

3.5 Plane conics up to aﬃne equivalence

Having dispensed with the classification of degree one polynomials up to affine equivalence in §3.3, we now turn to the case of degree two polynomials. Unlike the classification of linear polynomials, this classification is quite sensitive to the field K over which we work. The classification over the real numbers R is as follows:

Theorem 3.5.1. Every polynomial f(x, y) ∈ R[x, y] of degree two is aﬃne equivalent (Deﬁnition 3.3.3) to exactly one of the following:

1. x2 + y2 − 1

2. x2 − y2 − 1

3. y − x2

4. xy

5. x2

6. x(x − 1)

7. x2 + y2

8. x2 + y2 + 1

9. x2 + 1

Over the complex numbers C, the classiﬁcation is simpler:

Theorem 3.5.2. Every polynomial f(x, y) ∈ C[x, y] of degree two is aﬃne equivalent to exactly one of the following:

1. x2 + y2 − 1

2. y − x2

3. xy

4. x2

5. x(x − 1) 3.5 Plane conics up to aﬃne equivalence 39

We are not going to prove these theorems, though we will prove their “projective analogues” in §5.6. The proofs are elementary: the basic technique, which will also be used in the proofs our projective analogues of these theorems, amounts to little more than the idea of “completing the square” familiar from calculus (or from the derivation of the Quadratic Formula, say). However, the proofs in the affine case are slightly more tedious, as there is more case-by-case analysis. The projective analogues of this theorem are also in many ways more important and fundamental, as they are related to the problem of classifying symmetric bilinear forms, which is important in linear algebra. From the classification of degree two polynomials, we can easily obtain the classification of the conics given by their zero loci:

Corollary 3.5.3. For every polynomial f(x, y) ∈ R[x, y] of degree two, the corresponding 2 plane curve Z(f) ⊆ R is either empty, or aﬃne equivalent to exactly one of the following: 1. the circle Z(x2 + y2 − 1)

2. the hyperbola Z(x2 − y2 − 1)

3. the parabola Z(y − x2)

4. the pair of intersecting lines Z(xy)

5. the “doubled” line Z(x2)

6. the pair of parallel lines Z(x(x − 1))

7. the single point Z(x2 + y2)

Proof. This will follows from Theorem 3.5.1 and Lemma 3.3.1 once we prove that no two of the non-empty plane curves on the list are affine equivalent. (A priori, it is quite conceivable that two inequivalent polynomials determine equivalent plane curves, and, indeed, this is the case—the two inequivalent polynomials x2 + y2 + 1 and x2 + 1 both determine the empty curve, but, what we are about to argue is that, in fact, this does not happen, except in the aforementioned “empty curves” case.) This can be done in various ways. Perhaps the most direct way is to rule out the equivalence of different curves on the list on elementary topological grounds: affine transformations are, in particular, homeomorphisms, and the point is that very few pairs of curves on the list are even homeomorphic. (The reader who doesn’t know what “homeomorphism” means should just take this as a precise way of making precise the obvious “differences” we observe such as the fact that the parabola and circle are “connected,” whereas the hyperbola “has two components.” Similarly, the circle is “bounded” (compact), while the parabola and hyperbola are not.) The exceptions are: the hyperbola is homeomorphic to the two parallel lines (both are just disjoint unions of two copies of the real line), and the parabola is homeomorphic to the “double” line (both are homeomorphic to the real line). But these pairs of plane curves are easily seen to be distinct (inequivalent) on the grounds that affine transformations take lines to lines. For example, one need only write down three non-colinear points on the parabola to know that it cannot be equivalent to the “double” line. 40 Chapter 3 Affine Algebraic Plane Curves

3.6 Invariants of aﬃne conics

The most difficult thing to do in a classification problem is often to prove that various things are not equivalent. To do this, one typically tries to attach some kind of invariant— a number, say—to the objects being classified in such a way that this invariant depends only on the equivalence class of the object. The relevant invariants for plane conics are the discriminant and the degeneracy. Consider a typical polynomial

f(x, y) = Ax2 + Bxy + Cy2 + Dx + Ey + F (3.2)

∗ of degree at most two with coefficients in a field K where 2 ∈ K (for example: K = Q, R, or C). A key point is to note that f(x, y) can de described in terms of matrix multiplication, as follows. We let M(f) be the symmetric 3 × 3 matrix with entries in K defined by

 AB/2 D/2   M(f) := B/2 CE/2 . (3.3) D/2 E/2 F

∗ (Here we need to assume 2 ∈ K .) Then f(x, y) and the matrix M(f) are related via the formula:

 AB/2 D/2 x     f(x, y) = (x, y, 1) B/2 CE/2 y . (3.4) D/2 E/2 F 1

We call the matrix M(f) the matrix associated to f. Given any symmetric 3 × 3 matrix M with entries in K, we can deﬁne a polynomial f(M) ∈ K[x, y] of degree at most two by setting

f(M) = (x, y, 1)M(x, y, 1)tr.

It is clear from formula (3.4) that f 7→ M(f) establishes a bijection from polynomials in x, y of degree at most two to symmetric 3 × 3 matrices with inverse M 7→ f(M).

Deﬁnition 3.6.1. For f as above, we deﬁne the discriminant D(f) ∈ K and the degeneracy ∆(f) ∈ K of f by

D(f) := B2 − 4AC ∆(f) := det M(f).

We say that f(x, y) is degenerate if ∆(f) = 0 and non-degenerate otherwise.

The behaviour of D(f) and ∆(f) under aﬃne transformations is as follows. 3.6 Invariants of aﬃne conics 41

Lemma 3.6.2. For a polynomial f(x, y) ∈ K[x, y] of degree at most two, an aﬃne 2 transformation [A, t] ∈ Aﬀ(K ), and λ ∈ K, we have

D(λf · [A, t]) = λ2(det A)2D(f) ∆(λf · [A, t]) = λ3(det A)2∆(f).

Proof. The most direct proof is to just compute everything explicitly. The behaviour of D and ∆ under multiplying by λ is easy to understand (multiplying every entry of a 3 × 3 matrix by λ changes its determinant by λ3), so it is enough to deal with the case λ = 1. Suppose ! ! a b s A = t = c d t and f is given by (3.2). Then

f · [A, t] = f(ax + by + s, cx + dy + t) = A0x2 + B0xy + C0y2 + D0x + E0y + F 0, where

A0 = Aa2 + Bac + Cc2 B0 = 2Aab + B(ad + bc) + 2Ccd C0 = Ab2 + Bbd + Cd2 D0 = 2Aas + B(at + sc) + 2Cct + Da + Ec E0 = 2Abs + B(bt + ds) + 2Cdt + Db + Ed F 0 = As2 + Bst + Ct2 + Ds + Et + F.

Now a certain amount of laborious (but elementary) calculation yields

D(f · [A, t]) = (B0)2 − 4A0C0 = (ad − bc)2(B2 − 4AC) = det(A)2D(f)  A0 B0/2 D0/2  0 0 0  ∆(f · [A, s]) = det B /2 C E /2 D0/2 E0/2 F 0  AB/2 D/2 2   = (ad − bc) det B/2 CE/2 D/2 E/2 F = (det A)2∆(f). 42 Chapter 3 Aﬃne Algebraic Plane Curves

A slightly more sophisticated treatment of ∆(f) using the matrix M(f) proceeds as follows. We calculate

ax + by + s   f · [A, t] = (ax + by + s, cx + dx + t, 1)M(f) cx + dy + t 1 a c 0 a b s x       = (x, y, 1) b d 0 M(f) c d t y s t 1 0 0 1 1 so we see that the matrix

 A0 B0/2 D0/2  0 0 0  M(f · [A, t]) = B /2 C E /2 D0/2 E0/2 F 0 corresponding to f · [A, s] is given by

tr a b s a b s     M(f · [A, t]) = c d t M(f) c d t . 0 0 1 0 0 1

The second formula of the lemma now follows from standard properties of determinants (invariance under transpose, determinant of a product is the product of the determinants) using the trivial calculation

a b s   det c d t = det A. 0 0 1

∗ Corollary 3.6.3. Let K be a field with 2 ∈ K and let f ∈ K[x, y] be a polynomial of degree 2. Then ∆(f) = 0 iff ∆(g) = 0 for some (equivalently any) g affine equivalent to f. Similarly, when K = R, the “sign” of Df in {+, −, 0} depends only on the affine equivalence class of f.

We can use D and ∆ to distinguish between most of the pairs of plane curves on the list in Theorem 3.5.1. The values of D and ∆ for the curves on that list are tabulated 3.7 Exercises 43 below: f(x, y) D(f) ∆(f) x2 + y2 − 1 (−) 6= 0 x2 − y2 − 1 (+) 6= 0 y − x2 (0) 6= 0 xy (+) 0 x2 (0) 0 x(x − 1) (0) 0 x2 + y2 (−) 0 x2 + y2 + 1 (−) 6= 0 x2 + 1 (0) 0 At this point, we shouldn’t be shy about admitting that the deﬁnitions of the discriminant and the ∆-invariant seem to “come out of the blue.” Later, in Proposition 5.5.3, we will give a completely “geometric” meaning of the discriminant.

3.7 Exercises

Exercise 3.1. Suppose K is an algebraically closed ﬁeld (take K = C if you want) and f ∈ K[x, y] is a non-constant polynomial. Prove that Z(f) 6= ∅.

Exercise 3.2. Suppose f ∈ K[x, y] is a non-constant polynomial and n is an integer greater than one. Prove that every point of Z(f n) is a singular point. More generally, prove that if f = gh with g and h non-constant, then any point of Z(g) ∩ Z(h) is a singular point of Z(f).

2 Exercise 3.3. Let f ∈ K[x, y] be a polynomial, [A, t] ∈ Aff(K ) an affine transformation. Prove that a point P ∈ Z(f) is a singular point of Z(f) iff [A, t]−1(P ) is a singular point of Z(f · [A, t]).

Exercise 3.4. Prove that being affine equivalent (Definition 3.3.3) is an equivalence relation on K[x, y]. Exercise 3.5. Consider the degree two polynomials f := x2 + y2, g := −x2 − y2 in 2 R[x, y]. Show that there is no A ∈ Aff(R ) such that g = f · A. (Nevertheless, we do consider f and g to be affine equivalent because (−1)f = g.)

Exercise 3.6. For each of the following polynomials f ∈ R[x, y] decide which of the ﬁve polynomials in the list of Theorem 3.5.1 is aﬃne equivalent to f:

1. zy − x2

2. xy

3. x2 + xy + z2

4. 2x2 + 2xz + z2 + 2xy + y2 44 Chapter 3 Aﬃne Algebraic Plane Curves

2 Exercise 3.7. For f ∈ R[x, y], let Stab f be the set of A ∈ Aﬀ(R ) such that f · A = f. 2 −1 1. Show that Stab f is a subgroup of Aﬀ(R )—i.e. show that Id ∈ Stab f, that A ∈ Stab f whenever A ∈ Stab f, and that AA0 ∈ Stab f whenever A, A0 ∈ Stab f.

2. Note that Stab f = Stab(λf) for any non-zero λ ∈ R. Show that if f and g are 2 Aff-equivalent, then Stab f is conjugate to Stab g in Aff(R )—i.e. there is some 2 A ∈ Aff(R ) such that

Stab g = {A−1A0A : A0 ∈ Stab f}.

3. Show that Stab(x2 + y2) = Stab(x2 + y2 − 1) is equal to the orthogonal group 2 O2 < Aﬀ(R ). 4. Calculate Stab f for f = y − x2. Do you recognize this group? (Forget about the 2 description of it as a subgroup of Aﬀ(R ).) Chapter 4

Projective Spaces

n In this brief chapter we shall introduce and study the projective spaces KP for a field n n K. The idea of the construction of KP is to enlarge K by adding some additional n “points at infinity,” one for each line through the origin in K . One surprising feature of the construction is that the newly added points “at infinity” are actually “on the same footing” as the old “finite” points. We shall make this precise in §4.4 by introducing and studying the projective general linear groups PGLn(K). The group PGLn(K) acts on n n KP through projective transformations in a manner similar to the way Aff(K ) acts on n n K . Indeed, we shall see that Aff(K ) can be identified with the subgroup of PGLn(K) n n consisting of those A ∈ PGLn(K) such that the associated map A : KP → KP takes n n the locus of “finite points” K ⊆ KP into itself. In §4.2 we discuss linear subspaces of projective spaces—this concept leads to the definition of a line and to notions of “general position” which we be important later. n Special attention will be given to the real projective spaces RP and the complex n projective spaces CP in §4.3. A full appreciation of this section requires some background in topology—this material is not strictly necessary elsewhere in the text.

4.1 Motivation and construction

n Before we give the general definition of the projective spaces KP , let us motivate the 1 construction by discussing RP . One “problem” with the real line R is that it is not compact. The general notion of compactness is a concept of topology with which we shall not particularly concern ourselves here, but in this case, let us rephrase the issue a bit by noting that there are sequences of points a1, a2,... in R which do not have a convergent subsequence. For example, 1, 2, 3,... is such a sequence. Now, the astute reader with some calculus background would probably suggest that the limit of this sequence should be “∞,” so let us try to make some sense of this. We are led to attempt to enlarge the real line R by 1 ` adding one point ∞, thus obtaining a (slightly) larger space RP = R {∞}. But now 1 we have to decide how to “topologize” this larger set RP . In other words, we have to 1 give some precise definition of when a sequence of points a1, a2,... in RP “converges to ∞.” A very reasonable way to do this is to agree that a1, a2,... converges to ∞ iff the

45 46 Chapter 4 Projective Spaces

sequence 1/a1, 1/a2,... converges to 0 (in the usual epsilon-delta sense), where we agree that 1/∞ := 0. 1 With this understanding, let us prove that any sequence a1, a2,... in RP has a convergent subsequence. First of all, if infinitely many of the ai are equal to ∞, then those ai of course give a subsequence converging to ∞. On the other hand, if only finitely many of the ai are equal to ∞, then we discard them to obtain a subsequence contained 1 in R ⊆ RP . We thus reduce to the case where our sequence a1, a2,... is contained in 1 R ⊆ RP . Now, if the magnitudes |ai| of the ai are bounded, say by M, then we can find a convergent subsequence by basic results from calculus or topology (because the closed interval [−M,M] is compact). On the other hand, if the magnitudes |ai| are unbounded, then we can find an increasing sequence of integers

0 < i1 < i2 < ···

so that |aij | > j for each j ∈ {1, 2,... }. The subsequence ai1 , ai2 ,... of a1, a2,... then

converges to ∞ because |1/aij | < 1/j for each j ∈ {1, 2,... }. 1 There are many other ways to think about this construction of RP . For example, we 1 can view RP as being obtained from two copies R1, R2 of the real line R by identifying a non-zero point x ∈ R1 with the (non-zero) point 1/x ∈ R2. Every point in the resulting “glued up” space will have a unique representative in R1, with the exception of the origin 0 ∈ R2, which plays the role of ∞ in the previous construction. 1 Another way to think about the construction of RP more in line with what we will do in general, is to notice that any point x ∈ R determines a line Lx passing through the 2 origin in R —namely, the line with slope x. Notice that every line through the origin 2 in R will be equal to Lx for a unique real number x with one exception: the vertical line L∞ through the origin, which we might reasonably describe as the “line through the origin with slope ∞.” The reader with some background in linear algebra will no 2 doubt realize that a line through the origin in R is the same thing as a one dimensional 2 subspace of the vector space R . This is the point of view that will generalize well. 1 Yet another way to think about RP , also in line with what we will do later, is to think of a real number x as a ratio of two real numbers r = x/1 = 2x/2 = ··· where the denominator is non-zero. With this understanding, we might reasonable agree that ∞ can also be viewed as a ratio of two real numbers: ∞ = 1/0 = 5/0 = ··· , where the numerator is non-zero. To make this a bit more formal, we can consider the set 2 R \{(0, 0)} of pairs of real numbers (x, y), not both zero, and impose the equivalence relation ∼ where we declare (x1, y1) ∼ (x2, y2) iff x1y2 = y1x2. Notice that, if y1 and y2 are non-zero, we have (x1, y1) ∼ (x2, y2) iff the real numbers x1/y1 and x2/y2 are equal. Denote the equivalence class of (x, y) by [x : y]. Notice that every such equivalence class is of the form [x : 1] for a unique real number x with one exception: the equivalence class [1 : 0] (which is equal to the equivalence class [x : 0] for any non-zero real number x). With all this motivation, we are led to the general definition of the projective spaces n KP : n Definition 4.1.1. For a field K, we let KP denote the set of one dimensional subspaces n+1 n of the K-vector space K . We call KP the n-dimensional projective space over K. n+1 n For a non-zero vector x ∈ K \{0}, we let [x] ∈ KP denote the span of x. If 4.1 Motivation and construction 47

n+1 x = (x0, . . . , xn) ∈ K , we write [x0 : ··· : xn] as an alternative to [x]. More abstractly, if V is a K-vector space, then we let P(V ) denote the set of one dimensional subspaces of V . We call P(V ) the projectivization of V . n n+1 Notice that every point of KP is of the form [x] for some x ∈ K \{0} (because any one dimensional subspace of a vector space is spanned by any non-zero element of ∗ n it) and we have [x] = [y] iff x = λy for some λ ∈ K . We can thus view KP as the set n+1 ∗ of equivalence classes in K \{0} where we declare x ∼ y iff x = λy for some λ ∈ K . With this understanding, we can also view [x] as denoting the ∼-equivalence class of n+1 x ∈ K \{0}. ∗ If (x0, . . . , xn) = λ(y0, . . . , yn) for λ ∈ K , then clearly xi = 0 iff yi = 0, so the n+1 question of whether xi = 0 for a point x = (x0, . . . , xn) ∈ K \{0} depends only on the equivalence class [x] of x. Hence we can make the following: n Definition 4.1.2. Let U0,...,Un ⊆ KP be the subsets defined by n Ui := {[x0 : ··· : xn] ∈ KP : xi 6= 0}. n n The subset Un of KP will be called the set of finite points of KP , and its complement n n KP \ Un = {[x0 : ··· : xn] ∈ KP : xi = 0} n will be called the set of infinite points of KP . n Note that KP = U0 ∪ · · · ∪ Un. n n Definition 4.1.3. Define a bijection φi : K → Ui ⊆ KP by n n φi : K → KP (a1, . . . , an) 7→ [a1 : ··· : ai−1 : 1 : ai : ··· : an]. −1 The inverse φi of φi is given by −1 φi [x0 : ··· : xn] = (x0/xi, . . . , xi−1/xi, xi+1/xi, . . . , xn/xi). n−1 n Similarly, we let ψi : KP → KP \ Ui be the bijection defined by n−1 n ψi : KP → KP \ Ui [x0 : ··· : xn−1] 7→ [x0 : ··· : xi−1 : 0 : xi : ··· : xn−1]. n n In particular, φn gives a bijection between K and the set Un of finite points of KP . n n We often suppress notation for this bijection and simply regard K as a subset of KP n by identifying K with Un via φn. In other words, we don’t make much distinction n between a point (x1, . . . , xn) ∈ K and the “corresponding” point [x1 : ··· : xn : 1] of n n−1 n Un ⊆ KP . Similarly, we identify KP with the set of infinite points in KP via the bijection ψi, so if there is no chance of confusion we do not make any distinction between n−1 n [x0 : ··· : xn−1] ∈ KP and [x0 : ··· : xn−1 : 0] ∈ KP . In particular, notice that we n have identified the set of points at infinity in KP with the set of lines through the origin n n in (one dimensional linear subspaces of) K = Un ⊆ KP . 1 For example, in the case n = 1, the “finite points” of KP are those of the form [x : y] with y 6= 0, and there is a single infinite point [1 : 0], often denoted ∞, corresponding to 0 the single point of KP . 48 Chapter 4 Projective Spaces

4.2 Linear subspaces of projective space

The definition of P(V ) for an abstract K-vector space V is convenient, even if one is n only interested in KP , for essentially the same reasons that it is convenient to have an n abstract notion of “vector space” even if one is only interested in the vector space K —for example, because “abstract” vector spaces are going to arise naturally as subspaces of n n n+1 K . By definition, we have KP = P(K ). (The notation is unfortunate.) Notice also that if V is a subspace of a vector space W , then P(V ) is a subset of P(W ) because any one dimensional subspace of V is also a one dimensional subspace of W . In particular, a n+1 n+1 subspace V of the vector space K (a “linear subspace of K ”) gives rise to a subset n P(V ) ⊆ KP . n n+1 Definition 4.2.1. A linear subspace of KP of dimension d is a subsset of KP of n+1 the form P(V ), for V a linear subspace of K of dimension d + 1 (as a K vector space). A linear subspace of dimension one (resp. two) is often called a line (resp. plane), n while a linear subspace of KP of dimension n − 1 (“codimension one”) is often called a hyperplane.

n For example, the set of “infinite” points of KP is a hyperplane—it is P(V ) where V n+1 is the subspace of K spanned by the first n basis vectors e0, . . . , en−1. n Definition 4.2.2. Points P1,...,Pm ∈ KP are said to be in general position iff, for all n k ∈ {1,..., max m, n + 1}, no k of the Pi are contained in a linear subspace of KP of dimension less than k − 1.

n For example, when n ≥ 2, three points P1,P2,P3 ∈ KP are in general position iﬀ 1 they are not contained in any line. Three points P1,P2,P3 ∈ KP are in general position iﬀ they are distinct.

4.3 Real and complex projective spaces

n In order to get a feeling for the general projective spaces KP introduced in §4.1 it helps to think about the cases where K = R and K = C. n n+1 By deﬁnition, RP is the set of line through the origin in R . Each such line L will intersect the n-dimensional sphere

n n+1 S := {x ∈ R : |x| = 1} in precisely two points, which are “antipodal” in the sense that one is obtained from the other by multiplying the coordinates by −1. Indeed, if we choose any non-zero point x ∈ L, then the two points in question are x/|x| and −x/|x|. “Conversely,” if we have a n n+1 point x ∈ S ⊆ R , then we get a line through the origin Lx by considering the span n of x. The line Lx intersects S at the points x and the antipodal point −x. We thus see n n that RP can be obtained from the n-dimensional sphere S by identifying antipodal points. Alternatively, we can consider the upper hemisphere

n n S+ := {x = (x0, . . . , xn) ∈ S : xn > 0} 4.3 Real and complex projective spaces 49 of Sn. Notice that projecting onto the ﬁrst n coordinates deﬁnes a (continuous) bijection n from S+ to the n-dimensional disc

n n D := {x = (x0, . . . , xn−1) ∈ R : |x| ≤ 1}. The inverse of this bijection is given by

p 2 x = (x0, . . . , xn−1) 7→ (x0, . . . , xn−1, 1 − |x| ).

n+1 By reasoning as above, we see that each line through the origin in R will intersect n the upper hemisphere S+ in either one point or two points; this intersection will consist n n+1 of two antipodal points iﬀ the line is contained in the hyperplane R ⊆ R where the n last coordinate xn is zero. From this perspective we see that RP can be obtained from the n-dimensional disc Dn by identifying antipodal points of its boundary sphere Sn−1. 1 1 For example, RP is the quotient of S by the antipodal map x 7→ −x, but it can also be described by identifying the two boundary points of the upper half circle (which is homeomorphic to a closed interval), so it is also a circle. The quotient map 1 1 1 2 1 S → RP = S can be viewed as the map z 7→ z if we view the circle S as the set of complex numbers z of magnitude one. 2 2 3 The set RP can similarly be viewed as the quotient of the usual sphere S ⊆ R 2 2 obtained by identifying antipodal points, or as the quotient of the unit disc D ⊆ R obtained by identifying antipodal points on the boundary circle S1 ⊆ D2. n n The story for complex projective space CP is similar. For any point [z] ∈ CP , we can ﬁnd a representative z ∈ [z] lying in the sphere

2n+1 n+1 S := {z ∈ C : |z| = 1}.

0 2n+1 0 n 0 Two points z, z ∈ S determine the same point [z] = [z ] of CP iﬀ z = λz for some λ ∈ C—in fact, taking | |, we see that |λ| = 1, so λ will lie in the unit circle 1 S = {λ ∈ C : |λ| = 1}.

n 2n+1 1 This shows that CP can be described as the quotient of S by the free action of S given by (λ, z) 7→ λz. For the student with some background in topology, we remark that all of the descrip- n n tions of RP and CP given in these notes can actually be viewed as descriptions of a topological space (not just a set). In fact all of these descriptions are describing the same topology:

n n Definition 4.3.1. We view RP as a topological space by declaring a subset U ⊆ RP to be open iff the following equivalent conditions are satisfied:

1. The preimage of U under the map

n+1 n R \{0} → RP

n+1 given by x 7→ [x] is open in R . n n n 2. The preimage of U under the map S → RP given by x 7→ [x] is open in S . 50 Chapter 4 Projective Spaces

n n 3. The preimage of U under the map S+ → RP given by x 7→ [x] = Span x is open n in S+. n n 4. The preimage of U ∩ Ui under the bijection φi : R → Ui ⊆ RP (Deﬁnition 4.1.3) is open for each i ∈ {0, . . . , n}.

(The equivalence of these conditions is Exercise 4.2.)

n n Definition 4.3.2. We topologize CP by declaring a subset U ⊆ CP to be open iff the following equivalent conditions are satisfied:

1. The preimage of U under the map

n+1 n C \{0} → CP

n+1 given by z 7→ [z] is open in C . 2n+1 n 2n+1 2. The preimage of U under the map S → CP given by z 7→ [z] is open in S . n n 3. The preimage of U ∩ Ui under the bijection φi : C → Ui ⊆ CP is open for each i ∈ {0, . . . , n}.

n n In particular, CP and RP are both quotients of spheres and spheres are compact, so we have:

n Proposition 4.3.3. For each n, real projective space RP and compact projective space n CP are compact topological spaces. 1 2 In Exercise 4.4 you will show that CP is homeomorphic to the 2-sphere S , so that 3 1 the map S → CP given by z 7→ [z] discussed above (the quotient of the circle action on S3) can be viewed as a map f : S3 → S2 for which f −1(x) is a circle for each x ∈ S2. The map f is called the Hopf ﬁbration—it plays a role in various branches of topology. It is the most basic example of an “interesting” map from a sphere to a sphere of lower dimension.

4.4 Projective transformations

Now we will deﬁne the projective analogue of the aﬃne transformations studied in §1.4. First, notice that an invertible linear transformation A ∈ GLn+1(K) gives rise to a bijection

n n A : KP → KP (4.1) [x] 7→ [Ax].

∗ This is well-deﬁned because if x = λy for λ ∈ K , then λAx = Aλx = Ay, so [Ay] = [Ax]. We also need to know that A is invertible to ensure that Ax 6= 0 when x 6= 0, so that [Ax] n makes sense as a point of KP . The bijections (4.1) are called projective transformations. Clearly the composition A ◦ B of the projective transformations associated to invertible matrices A, B ∈ GLn+1(K) is the projective transformation AB associated to their 4.4 Projective transformations 51 product. In particular, the inverse of the projective transformation A associated to −1 A ∈ GLn+1(K) is the projective transformation A associated to its inverse. Notice ∗ that for λ ∈ K and A ∈ GLn+1(K), we have A = λA (for the same reason that A is well-deﬁned). We are thus led to consider the following group:

Deﬁnition 4.4.1. The projective general linear group PGLn(K) is the quotient of ∗ GLn+1(K) by the (normal—in fact central) subgroup {λI : λ ∈ K }. We denote the image of A ∈ GLn+1(K) in PGLn(K) by A.

Explicitly, two invertible matrices A, B ∈ GLn+1(K) have the same image in PGLn(K) ∗ iﬀ A = λB for some λ ∈ K .

Remark 4.4.2. The indexing of PGLn(K) varies. Some people might write PGLn+1(K) for what I have called PGLn(K).

The next result justiﬁes our use of the notation A for both the image of a matrix n A ∈ GLn+1(K) in PGLn(K) and for the associated projective transformation A : KP → n KP .

Proposition 4.4.3. Two invertible matrices A, B ∈ GLn+1(K) have the same image in n n PGLn(K) iﬀ the projective transformations A, B : KP → KP are equal.

Proof. By deﬁnition of PGLn(K), if A and B have the same image in PGLn(K), then ∗ we can write A = λB for some λ ∈ K and we already noted above that this implies that the projective transformations A and B are equal. n n Conversely, suppose the projective transformations A, B : KP → KP are equal. ∗ Then, since A[ei] = B[ei] for i = 0, . . . , n, we can write Aei = λiBei for some λi ∈ K . Also, since A[1 : ··· : 1] = B[1 : ··· : 1], we can write

Ae0 + ··· + Aen = λ(Be0 + ··· + Ben)

∗ for some λ ∈ K . Putting these together we see that

λ0Be0 + ··· + λnBen = λBe0 + ··· + λBen.

n+1 But B is invertible, so Be0, . . . , Ben is a basis for K , hence the above equality implies that λ = λ1 = ··· = λn and we conclude that A = λB, so A and B have the same image in PGLn(K).

Theorem 4.4.4. The subgroup

n n G := {A ∈ PGLn+1(K): A(K ) = K } of PGLn+1(K) consisting of projective transformations that “preserve the locus of finite n n n n points K ⊆ KP ” is isomorphic to the group Aff(K ) of affine transformations of K n via the restriction map A 7→ A|K . 52 Chapter 4 Projective Spaces

n n Proof. Suppose A ∈ G, so the projective transformation A : KP → KP takes the locus n n n n K ⊆ KP of finite points into itself. Equivalently, since A : KP → KP is bijective, n−1 n this means that A takes the locus KP ⊆ KP of infinite points into itself. Pick A ∈ GLn+1(K) mapping to A in PGLn(K). Since [e0],..., [en−1] are infinite points of n KP , the points A[ei] = [Aei] must also be infinite points for i = 0, . . . , n − 1, which means that the last entry of each vector Aei (i = 0, . . . , n − 1) must be a zero. On the other hand, A must be invertible, so the last entry λ of Aen therefore cannot be zero. −1 Then L(A) := λ A ∈ GLn+1(K) also maps to A in PGLn(K) and is the unique such matrix with lower right entry equal to 1. Note that the (n + 1) × (n + 1) matrix L(A) takes the “block upper triangular” form ! B t L(A) = , (4.2) 0 1 where B is an n × n matrix and t is a column vector with n rows. A matrix in this block upper triangular form is invertible iff B is invertible (for example, because one can check that det L(A) = det B for such a “block upper triangular” matrix L(A)). From the definition of matrix multiplication we see that two such block upper triangular matrices multiply according to the rule ! ! ! B t B0 t0 BB0 Bt0 + t = . (4.3) 0 1 0 1 0 1

In particular, this shows that L(A1A2) = L(A1)L(A2), so that L may be viewed as an 0 isomorphism of groups from G to the subgroup G of GLn+1(K) consisting of invertible matrices of the “block upper triangular form” described above. Furthermore, if we compare the multiplication formula (4.3) to the formula (1.3) for composition of aﬃne transformations, we see that ! B t 7→ [B, t] 0 1

0 n is an isomorphism from G to Aff(K ). To see that the isomorphism A 7→ [B, t] from G n to Aff(K ) just described is given by the indicated restriction map, just notice that if we n n fix a point x ∈ K , then the corresponding point of KP is [x : 1] and we have

A[x : 1] = [L(A)(x, 1)] " ! !# B t x = 0 1 1 " !# Bx + t = , 1

n n which is nothing but [B, t](x) = Bx + t ∈ K viewed as a point of KP in the usual manner. 4.5 Exercises 53

We now prove the analogue of Proposition 1.4.3 for projective transformations.

n Proposition 4.4.5. If P0,...,Pn+1 are n + 2 points of KP in general position (Deﬁni- tion 4.2.2), then there is a unique A ∈ PGLn(K) satisfying A[ei] = P for i = 0, . . . , n and A[1 : ··· : 1] = Pn+1.

n+1 Proof. Choose representatives x0, . . . , xn+1 ∈ K \{0} for the Pi. Since the Pi are in n+1 general position, x0, . . . , xn must be a basis for K , so we can write n X xn+1 = λixi i=0 for unique λi ∈ K. Each λi must be non-zero, otherwise xn+1 would be in the span of a proper subset of {x0, . . . , xn}, in violation of the general position assumption. The image th A ∈ PGLn+1(K) of the matrix A ∈ GLn+1(K) whose i column is λixi is as desired.

4.5 Exercises

2 Exercise 4.1. In Definition 4.1.2 we defined subsets U0,U1,U2 of KP , and, more n 2 2 generally, U0,...,Un of KP . In Definition 4.1.3 we defined bijections φi : K → Ui ⊆ KP , n n and, more generally, φi : K → Ui ⊆ KP . −1 −1 −1 2 a) Describe the subsets φ1 (U1 ∩ U2), φ1 (U0 ∩ U1 ∩ U2), and φ2 (U1 ∩ U2) of K . −1 −1 b) We obtain a bijection φ12 from φ1 (U1 ∩ U2) to φ2 (U1 ∩ U2) by mapping a point −1 in φ1 (U1 ∩ U2) into U1 ∩ U2 using the map φ1, then taking the inverse of the resulting point under φ2. Give an explicit formula for φ12.

n −1 c) More generally, for an arbitrary projective space KP , describe the subsets φi (Ui ∩ −1 −1 Uj) and the bijections φij : φi (Ui ∩ Uj) → φj (Ui ∩ Uj) deﬁned in the analogous manner.

The next three exercises probably require some background in topology.

n Exercise 4.2. Show that, for a subset U ⊆ RP , the conditions of Definition 4.3.1 are equivalent. Also show that the conditions of Definition 4.3.2 are equivalent for a subset n U ⊆ CP . n Exercise 4.3. For K = R or C, prove that the subsets Ui ⊆ KP defined in Defini- n tion 4.1.2 are open in the topology on KP defined in Definitions 4.3.1 and 4.3.2. (If you haven’t done the previous exercise, then just use, say, the first condition in each n of those definitions.) Prove that the bijections φi : K → Ui of Definition 4.1.3 are n homeomorphisms when Ui is given the subspace topology from the inclusion Ui ⊆ KP . 1 2 Exercise 4.4. Show that CP is homeomorphic to the 2-sphere S . n+1 Exercise 4.5. Let A ∈ GLn+1(K), x ∈ K \{0}. In terms of basic concepts from n linear algebra, what does it mean to say that [x] ∈ KP is a fixed point of the projective n n transformation A : KP → RP associated to A? 54 Chapter 4 Projective Spaces

2 Exercise 4.6. Let L1,L2,L3 be three lines in R with empty intersection: L1∩L2∩L3 = ∅. 2 2 How many connected components does R \ (L1 ∪ L2 ∪ L3) have? What happens if R is 2 replaced by RP ? Hint: Move one of the lines to inﬁnity by a projective transformation. (In fact you can move them even “more freely” as described in Exercise 4.10.)

There are an inﬁnite number of possible variations on Proposition 4.4.5. Here are some:

3 3 Exercise 4.7. Suppose L and M are lines in KP and P is a point of KP . Prove that 3 there is a line in KP containing P and intersecting both L and M. n Exercise 4.8. Suppose P(V ) ⊆ KP is a linear subspace. Prove that every projective transformation of P(V ) (invertible linear transformation V → V , up to rescaling) extends n to a projective transformation of KP . 2 Exercise 4.9. Let L be a line in KP , P a point of L, M a line not containing P . Prove 2 2 that there is a projective transformation A : KP → KP taking L to the line at inﬁnity, P to the point [0 : 1 : 0], and M to the line Z(y) = P(Span(e0, e1)). 2 Exercise 4.10. Suppose L1,...L4 are four lines in KP such that the intersection of 0 0 2 any three is empty. Suppose L1,...,L4 are four lines in KP with the same property. 2 2 0 Prove that there is a unique projective transformation A : KP → KP taking Li onto Li for i = 1,..., 4. Hint: This statement is “dual” to the statement of Proposition 4.4.5 in the case n = 2. Chapter 5

Projective Plane Curves

In this chapter, we will specialize our study of projective spaces in Chapter 4 to the case 2 2 of the projective plane KP over a ﬁeld K. Recall that we identify the aﬃne plane K over K with the subset

2 U2 = {[x : y : z] ∈ KP : z 6= 0}

2 2 of the projective plane KP via the map (x, y) 7→ [x : y : 1], thus we view K as a subset 2 of KP , which we call the set of finite points. The complementary subset of infinite points 1 is called the line at infinity; it is identified with KP via [x : y : 0] ↔ [x : y]. We will see 2 that the projective plane has many nice features that are lacking in the affine plane K . 2 2 For example, in K , two distinct lines may or may not intersect, whereas in KP , any two distinct lines intersect in precisely one point (§5.4). Just as a polynomial in two variables f ∈ K[x, y] gives rise to an affine algebraic 2 plane curve Z(f) ⊆ K , we shall see in §5.1 that a homogeneous polynomial (also called a 2 form) in three variables f ∈ K[x, y, z] gives rise to a projective plane curve Z(f) ⊆ KP . Using this construction, we shall revisit some of the geometry we studied in Chapter 3. In §5.8 we give a brief treatment of intersection multiplicity and Bezout’s Theorem, to the extent that we shall need it elsewhere. This section is rather tedious and not a particularly satisfactory treatment of intersection multiplicity, as my hands were tied by my refusal to make use of any concepts from commutative algebra (besides the notion of fields and polynomials). In §5.5 we see that the discriminant of a real affine conic, introduced in §3.6 has a natural “geometric” interpretation in projective geometry. In §5.6 we carry out the classification of projective conics (over R and C) promised in §3.5. The remaining sections of this chapter are largely independent of one another and consist of a hodge-podge of topics in the geometry of projective plane curves. In §5.7, we discuss the cross ratio, which may be thought of as a classification of ordered lists of four 1 points in KP , up to projective equivalence. In §5.9 we prove some “classical” theorems of projective geometry. The theorem of Desargues proved there plays an important role in axiomatic projective geometry. In §5.10 we show that the set of non-singular points of (certain) smooth cubic curves can be given the structure of an abelian group. We describe this group explicitly in the case of the “cuspital cubic” in §5.11.

55 56 Chapter 5 Projective Plane Curves

5.1 Homogeneous polynomials and zero loci

A polynomial f(x, y, z) ∈ K[x, y, z] is called homogeneous of degree d (or a form of degree d) iﬀ it can be written in the form

X i j k f(x, y, z) = Aijkx y z i+j+k=d for coeﬃcients Aijk ∈ K. Usually we assume that at least one of the Aijk is non-zero, so that f 6= 0, though in some situations it is convenient to have the convention that the zero polynomial is homogeneous of degree d for every d. A form of degree one (resp. two, three, . . . ) is often called a linear (resp. quadratic, cubic,...) form. If f is a form of degree d then for any λ ∈ K we have X f(λx, λy, λz) = (λx)i(λy)j(λz)k (5.1) i+j+k=d X = λi+j+kxiyjzk i+j+k=d = λdf(x, y, z).

To any degree d polynomial

X i j f(x, y) = Aijx y i+j≤d in two variables we associate a homogeneous polynomial f˜ in three variables, called the homogenization of f(x, y), by setting

X i j d−i−j f˜(x, y, z) := Aijx y z . i+j≤d

This association f 7→ f˜ yields a bijection between degree d polynomials in x, y and homogeneous degree d polynomials in x, y, z whose inverse is given by f(x, y, z) 7→ f(x, y, 1). For any homogeneous polynomial f(x, y, z), we can consider the subset

2 Z(f) := {[x : y : z] ∈ KP : f(x, y, z) = 0} (5.2) 2 of the projective plane KP . This makes sense because the question of whether f(x, y, z) = 3 0 depends only on the equivalence class [x : y : z] of (x, y, z) ∈ K \{0}, in light of the formula (5.1) (in the case of a non-zero λ).

Definition 5.1.1. For a homogeneous polynomial (form) f ∈ K[x, y, z] (usually assumed 2 to be of positive degree), the subset Z(f) ⊆ KP defined by (5.2) is called the zero 2 locus of f. A subset S ⊆ KP equal to Z(f) for some homogeneous polynomial f (of positive degree) is called a projective plane curve and f is called a defining equation 2 for S. If f ∈ K[x, y] is any non-constant polynomial, the zero locus Z(f˜) ⊆ KP of the 2 homogenization of f is called the projective completion of Z(f) ⊆ K . 5.2 Smooth and singular points 57

The usual remarks are in order: What we call the “projective completion of Z(f)” really depends highly on the polynomial f deﬁning Z(f)—it is not really intrinsic to the 2 subset Z(f) ⊆ K . Notice that, for a polynomial f ∈ K[x, y], we have f˜(x, y, 1) = f(x, y),

2 2 2 2 so (x, y) ∈ K is in Z(f) ⊆ K iﬀ [x : y : 1] ∈ KP is in Z(f˜) ⊆ KP . In other words, the 2 2 ﬁnite points of the projective completion Z(f˜) ⊆ KP of Z(f) ⊆ K are just the points of Z(f):

2 Z(f˜) ∩ K = Z(f).

If f ∈ K[x, y, z] is homogeneous, notice that f(x, y, 0) ∈ K[x, y] is homogeneous (though 1 2 possibly zero even if f is non-zero)—the intersection Z(f) ∩ KP of Z(f) ⊆ KP with the 1 2 line at inﬁnity KP ⊆ KP is given by 1 1 Z(f) ∩ KP = {[x : y] ∈ KP : f(x, y, 0) = 0}. This intersection with the line at inﬁnity is often of interest even in the case where f = g˜ is the homogenization of a polynomial g ∈ K[x, y]—it encodes, in an appealing geometric 2 way, the “asymptotic” behaviour of g(x, y) for “large” (x, y) ∈ K . We shall return to this point in §5.5.

Remark 5.1.2. Given a homogeneous polynomial f ∈ K[x, y, z], the notation Z(f), used in isolation could possibly be confusing, because it might be understood as the projective 2 plane curve Z(f) ⊆ KP , or just as the “affine” zero locus 3 Z(f) = {(x, y, z) ∈ K : f(x, y, z) = 0} of f. In these notes, we always have the first meaning of Z(f) in mind. These two “zero loci” are not unrelated: The homogeneity of f and (5.1) show that the “affine” zero locus of f—let us call it C(Z(f)) for the moment, for clarity—is “invariant under rescaling,” meaning: If (x, y, z) ∈ C(Z(f)), then λ(x, y, z) ∈ C(Z(f)) for any λ ∈ K (and conversely ∗ 3 2 if λ ∈ K ). Note also that the projection map K \{0} → KP given by x 7→ [x] takes 2 C(Z(f)) (minus the origin) onto Z(f) ⊆ KP . For these reasons, the affine zero locus 2 C(Z(f)) is often called the cone on the projective zero locus Z(f) ⊆ KP .

5.2 Smooth and singular points

In this section we give the projective analogues of the deﬁnitions from §3.2.

2 Definition 5.2.1. For a homogeneous polynomial f ∈ K[x, y, z], a point P ∈ Z(f) ⊆ KP is called a non-singular (or smooth) point of Z(f) iff at least one of fx(P ), fy(P ), fz(P ) is non-zero. A point P ∈ Z(f) for which fx(P ) = fy(P ) = fz(P ) = 0 is called a singular point of Z(f). The subset of Z(f) consisting of smooth (resp. singular) points is denoted sm sing sing Z(f) (resp. Z(f) ). We say that f is non-singular or smooth iff Z(f) (K) is empty 1 for every field extension K ⊆ K. 1 sing It equivalent to ask that Z(f) (K) = ∅ for some algebraically closed field K containing K. 58 Chapter 5 Projective Plane Curves

As in the aﬃne case, the question of whether P ∈ Z(f) is a smooth point or a singular point depends highly on f, not just on the curve C = Z(f). The following lemma is often useful when trying to determine the singular points of a projective curve:

Lemma 5.2.2. For a homogeneous polynomial f ∈ K[x, y, z] of degree d, we have

df = xfx + yfy + zfz.

Consequently, if d is not zero in K (which holds if d > 0 and K = Q, R, C, or any 2 other ﬁeld of characteristic zero), then a point P ∈ KP is a singular point of Z(f) iﬀ fx(P ) = fy(P ) = fz(P ) = 0.

Proof. Exercise 5.11.

Deﬁnition 5.2.3. If P is a non-singular point of Z(f), we deﬁne the tangent line to Z(f) at P to be the line

TP Z(f) := Z(xfx(P ) + yfy(P ) + zfz(P )).

The previous lemma implies that P ∈ TP Z(f). The projective notions of smoothness, tangent lines, etc. are compatible in every imaginable sense with those from the aﬃne setting of §3.2. Some precise statements along these lines are given in Exercise 5.12.

5.3 Linear change of variables and projective equivalence

The notion of aﬃne equivalence introduced in §3.3 also has an analogue in the projective setting. Given a form (homogeneous polynomial) f ∈ K[x, y, z] = K[x] of degree d and an invertible matrix A ∈ GL3(K), we obtain a form f · A ∈ K[x] of degree d by setting

(f · A)(x) := f(Ax).

As in §3.3, we think of the vector x = (x, y, z) of variables as a column vector, and the arguments of our polynomials as column vectors.

Definition 5.3.1. Two homogeneous polynomials f, g ∈ K[x, y, z] = K[x] are called ∗ projectively equivalent iff there is an invertible matrix A ∈ GL3(K) and a λ ∈ K so that 2 f(x) = λg(Ax). Two subsets S, T ⊆ KP are called projectively equivalent iff there is an A ∈ GL3(K) such that A(S) = T . When g = f · A for an invertible matrix A, we often say that g is obtained from f by ∗ a linear change of variables. Similarly, if g = λf for λ ∈ K , we say that g is a rescaling of f. The formula (5.1) shows that if A ∈ GL3(K) and f ∈ K[x, y, z] is homogeneous of degree d, then f · (λA) = λdf · A. So as long as we consider our homogeneous polynomials only up to rescaling (as is reasonable if we are concerned with their zero loci), f · A depends only on the image A ∈ PGL2(K) of A in the projective general linear group. 5.4 Lines and linear functionals 59

As in the aﬃne setting, it is straightforward to check that projective equivalence is in fact an equivalence relation (Exercise 5.1). As in the aﬃne case (Lemma 3.3.1), one sees that when f(x) = λg(Ax), the corresponding projective plane curves are related by

Z(f) = A−1(Z(g)). (5.3)

In particular, projectively equivalent homogeneous polynomials have projectively equivalent zero loci. As in the aﬃne case, it is easy to classify homogeneous linear polynomials up to projective equivalence—we shall see in the next section that there is only one equivalence class, regardless of the ﬁeld K over which we work.

5.4 Lines and linear functionals

2 Recall that, in Definition 4.2.1, we defined a line to be a subset L of KP equal to P(V ) for 3 some two dimensional subspace V of the K vector space K . By linear algebra, every such V arises as the kernel of some surjective (equivalently: non-zero) linear transformation 3 2 (A, B, C): K → K, hence a line may equivalently be defined as a subset L ⊆ KP such that

2 L = {[x : y : z] ∈ KP : Ax + By + Cz = 0}

3 2 for some (A, B, C) 6= (0, 0, 0) in K . In other words, a line is the zero locus in KP of a linear form. As in Definition 5.1.1, we call “Ax + By + Cz” a defining equation for L. (Compare the affine version of a line, discussed in §3.3.) Recall from linear algebra that if V is a K-vector space, then a linear transformation V → K is called a linear functional on V . The set

∗ V := HomK(V, K) of linear functionals on V is itself a K-vector space (under “pointwise addition and scalar multiplication”), called the dual vector space of V . The next lemma says that 2∗ 3 ∗ 2 [f] 7→ P(Ker f) establishes a bijection from KP = P((K ) ) to the set of lines in KP . Lemma 5.4.1. The line with defining equation Ax + By + Cz coincides with the line 0 0 0 0 0 0 ∗ with defining equation A x + B y + C z iff (A, B, C) = λ(A ,B ,C ) for some λ ∈ K . In 3 other words, the kernel of a linear functional f : K → K coincides with the kernel of the 0 3 0 ∗ linear functional f : K → K iff f = λf for some λ ∈ K . 0 0 0 ∗ 0 Proof. If (A ,B ,C ) = λ(A, B, C) for some λ ∈ R , then Z(Ax + By + Cz) = Z(A x + B0y + C0z) because A0x + B0y + C0z = λ(Ax + By + Cz), so A0x + B0y + C0z = 0 iff Ax+By+Cz = 0. On the other hand, suppose that Z(Ax+By+Cz) = Z(A0x+B0y+C0z) 0 0 0 ∗ and let us show that (A ,B ,C ) = λ(A, B, C) for some λ ∈ R . After possibly permuting the three coordinates everywhere, we can assume that A =6 0. Then [−B : A : 0] ∈ 2 0 0 0 Z(Ax + By + Cz) ⊆ RP , so, since Z(Ax + By + Cz) = Z(A x + B y + C z), we have [−B : A : 0] ∈ Z(A0x + B0y + C0z), hence AB0 = BA0, or, equivalently, B0 = λB, where we set λ := A0/A. Similarly, since [−C : 0 : A] ∈ Z(Ax + By + Cz), we find that AC0 = CA0, hence C0 = λC, and then (A0,B0,C0) = λ(A, B, C) as desired. 60 Chapter 5 Projective Plane Curves

2 In other words, lines in KP are in bijective correspondence with points of “another” 2∗ projective plane KP (whose coordinates we will call A, B, C, rather than x, y, z to avoid confusion) via the map taking the line with deﬁning equation Ax + By + Cz to the point [A : B : C]. The following result sums up basically everything there is to know about lines:

2 Proposition 5.4.2. Let K be a ﬁeld, KP the projective plane over K. 1. For any line L = Z(Ax + By + Cz), any point P of L is a smooth point and L = TP L is the tangent line to L at P .

1 2 2. The “line at inﬁnity” KP ⊆ KP is a line. 3. Any line L, other than the line at inﬁnity, is the projective completion of a unique 2 line in K . 2 4. Any two distinct lines in KP intersect in exactly one point. 2 2 5. If L ⊆ KP is a line and A ∈ GL3(K), then the image A(L) ⊆ KP of L under the 2 2 projective transformation A : KP → KP is also a line.

6. For any line L, there is some A ∈ GL3(K) such that A(L) is the line at inﬁnity.

Proof. (1): The partial derivations of f = Ax+By +Cz are fx = A, fy = B, and fz = C, at least one of which is a non-zero element of K by definition of a line. The equality L = TP L is immediate from the definition of TP L (Definition 3.2.2). (2): Note that z = 0x + 0y + 1z is a defining equation for the line at infinity. 3 Alternatively, note that the line at infinity is the projectivization of the subspace of K spanned by the first two standard basis vectors e0, e1 (cf. §4.2). (3): If L = Z(Ax + By + Cz) is not the line at infinity, then by Lemma 5.4.1 and the description of the line at infinity as Z(z), we see that either A or B is non-zero—let us 2 assume A 6= 0 as the case B 6= 0 is similar. Then Z(x + (B/A)y + (C/A)) is a line in K whose projective completion is

Z(x + (B/A)y + (C/A)z) = Z(A(x + (B/A)y + (C/A)z)) = L.

0 0 0 0 0 2 If Z(A x + B y + C ) (with (A ,B ) 6= (0, 0)) is another line in K whose projective completion is L, then we have L = Z(A0x + B0y + C0z), hence (A0,B0,C0) = λ(A, B, C) ∗ for some λ ∈ K by Lemma 5.4.1. Then we have (λA)(x + (B/A)y + (C/A)z) = A0x + B0y + C0, and hence

Z(x + (B/A)y + (C/A)z) = Z(A0x + B0y + C0).

(4): Suppose L = Z(Ax + By + Cz) and L0 = Z(A0x + B0y + C0z) are two distinct 0 0 0 3 lines. Then (A, B, C), (A ,B ,C ) ∈ K are linearly independent by Lemma 5.4.1, so, by linear algebra, the matrix A whose rows are (A, B, C) and (A0,B0,C0) gives a surjective 5.5 Invariants of aﬃne conics revisited 61

3 2 (i.e. rank two) linear map A : K → K ; the kernel of such a map is a one dimensional 3 2 0 subspace of K which, viewed as a point of KP , is the intersection of L and L . (5): This follows from (5.3). 3 (6): Write L = P(V ) for a two dimensional linear subspace V ⊆ K . By linear 3 algebra, we can ﬁnd a basis v0, v1, v2 for K such that v0, v1 is a basis for V . By linear algebra there is a unique A ∈ GL3 with Avi = ei, i = 0, 1, 2. This A takes the two 3 2 2 dimensional linear subspace V bijectively onto Span(e0, e1) ⊆ K , hence A : KP → KP 1 takes L = P(V ) bijectively onto the line at inﬁnity KP = P(Span(e0, e1)).

5.5 Invariants of aﬃne conics revisited

The discriminant (Deﬁnition 3.6.1 in §3.6) of a degree two polynomial f(x, y) ∈ R[x, y] has an elegant geometric interpretation in terms of projective plane curves, which we will give momentarily after a brief lemma.

Lemma 5.5.1. For A, B, C ∈ R (not all zero), the subset 1 2 2 S := {[x : y] ∈ RP : Ax + Bxy + Cy = 0} of the real projective line is empty (resp. consists of a single one point, consists of two distinct points) if B2 − 4AC < 0 (resp. B2 − 4AC = 0, B2 − 4AC > 0).

1 Proof. First of all, the point ∞ = [1 : 0] ∈ RP is in S iff A = 0. Every point other than 1 ∞ in RP can be written as [x : 1] for a unique real numer x, and such a point is in S iff Ax2 + Bx + C = 0. Now we divide into cases. If A = 0, then ∞ ∈ S and [x : 1] is in S iff Bx + C = 0. The equation Bx + C = 0 has exactly one solution when B 6= 0 and no solutions when B = 0 (because then C 6= 0 since we assume A, B, C aren’t all zero). In the latter case, we have B2 − 4AC = B2 > 0, and S has two points (namely ∞, [−C/B : 1]), while in the former case, we have B2 − 4AC = 0 and S consists only of the point ∞. The lemma is hence correct in the case A = 0. If A 6= 0, then ∞ ∈/ S, and hence the points of S are just the real numbers x for which Ax2 + Bx + C = 0. Since A 6= 0, the number of such real numbers x is as stated in the lemma by a an exercise with the Quadratic Formula that should be familiar from high-school algebra. Remark 5.5.2. Over the complex numbers, the corresponding result says that S consists of a single point if B2 − 4AC = 0 and two points otherwise. The “single” point should be thought of as occurring with “multiplicity two.” Cf. Bezout’s Theorem (§5.8).

Now we can see that the discriminant of a degree two polynomial f(x, y) ∈ R[x, y] is 2 2 just a way of encoding how the projective completion Z(f˜) ⊆ RP of Z(f) ⊆ R intersects 1 2 the line at infinity RP ⊆ RP : Proposition 5.5.3. Let f = Ax2 + Bxy + Cy2 + Dx + Ey + F be a polynomial of degree at most two with real coefficients A, . . . , F and discriminant D(f) = B2 − 4AC. Then: 1. D(f) < 0 iff Z(f˜) is disjoint from the line at infinity, 62 Chapter 5 Projective Plane Curves

1 2. D(f) > 0 iﬀ Z(f˜) ∩ RP consists of two points, and 1 3. D(f) = 0 iﬀ Z(f˜) ∩ RP consists of a single point. Proof. Note that the homogenization of f is

f˜ = Ax2 + Bxy + Cy2 + Dxz + Eyz + F z2, ˜ so a typical point [x : y : 0] of the line at infinity is on ZP(f) iff f(x, y, 0) = 0 iff Ax2 + Bxy + Cy2 = 0. The result now follows from the previous lemma.

For a generalization of these ideas, see Exercise 5.2.

5.6 Classiﬁcation of projective conics

Just as we studied the problem of classifying affine plane curves and polynomials in two variables up to affine equivalence (§3.3), we can also study the problem of classifying projective plane curves and homogeneous polynomials in three variables up to projective equivalence (Definition 5.3.1). In this section, we will classify homogeneous degree two polynomials (with real or complex coefficients) and the corresponding projective plane curves (also called conics, as in the affine case) up to projective equivalence. The projective classification is simpler than the affine one, so we can give complete proofs in this section. As in the affine case, the classification of conics is highly sensitive to the field over which we work. The classification is “easier” over C than it is over R. Many steps in the classification make sense over an arbitrary field K where 2 = 1+1 ∈ ∗ K (i.e. a field whose characteristic is not equal to 2). Consider a quadratic form f = Ax2 + Bxy + Cy2 + Dxz + Eyz + F z2 (5.4) with coefficients A, . . . , F in such a field K. As in §3.5, we consider the symmetric 3 × 3 matrix

 AB/2 D/2   M(f) := B/2 CE/2 (5.5) D/2 E/2 F

∗ with entries in K—notice that we have to assume 2 ∈ K to be able to divide by 2 in deﬁning the oﬀ-diagonal entries of our matrix M(f). We have

f(x) = xtrM(f)x, (5.6) where x is our vector of variables (x, y, z), viewed as a column vector as usual. Notice that (5.6) is a little more natural looking than the analogous formula (3.4) we encountered in the aﬃne case. In particular, notice that for A ∈ GL3(K), the formula (5.6) gives f(Ax) = (Ax)trM(f)Ax = xtrAtrM(f)Ax, 5.6 Classiﬁcation of projective conics 63 thus we see that M(f) and M(f · A) are related by

M(f · A) = AtrM(f)A. (5.7)

Proposition 5.6.1. Let K be a ﬁeld. For a quadratic form f ∈ K[x, y, z], the following are equivalent:

1. We can write f = gh for linear forms g, h ∈ K[x, y, z], hence Z(f) = Z(g) ∪ Z(h) is a union of “two” (possibly equal) lines.

2. Z(f) contains a line.

∗ These equivalent conditions imply condition (3) below. If 2 ∈ K , then the following two conditions are equivalent:

3. There is a singular point in Z(f).

4. The matrix M(f) is singular (not invertible).

Proof. Obviously (1) implies (2). Conversely, suppose Z(f) contains a line L. By −1 Proposition 5.4.2(6) there is an invertible matrix A ∈ GL3(K) such that A (L) is the line at infinity, hence Z(f · A) contains the line at infinity by Formula 5.3. If f · A = gh factors as a product of two homogeneous linear polynomials, then f = (g · A−1)(h · A−1) also has such a factorization. We thus reduce to proving that if Z(f) contains the line at infinity, then f factors as a product of two homogeneous linear polynomials. Suppose Z(f) contains the line at infinity. Then Z(f) contains the three points [0 : 1 : 0], [1 : 0 : 0], and [1 : 1 : 1], so we have

f(0, 1, 0) = f(1, 0, 0) = f(1, 1, 0) = 0.

If we write f as in (5.4), then these equalities are equivalent to A = B = C = 0, in which case we have the desired factorization:

f = z(Dx + Ey + F z).

2 (1) implies (3): Since the intersection of two lines in KP is non-empty by Proposi- tion 5.4.2(4), this follows from Exercise 5.3. 3 (3) iﬀ (4): Notice that for x ∈ K , we have

2M(f)x = (fx(x), fy(x), fz(x)) (5.8)

(up to a transpose). By linear algebra, the matrix M(f) is singular iﬀ M(f)x = 0 for 3 some x ∈ K \{0}. From (5.8), we see that this is equivalent to saying there is some 2 [x] ∈ KP for which fx(x) = fy(x) = fz(x) = 0. ∗ Since 2 ∈ K this is equivalent to saying there is a singular point in Z(f) by Lemma 5.2.2. 64 Chapter 5 Projective Plane Curves

Deﬁnition 5.6.2. A homogeneous degree two polynomial f ∈ K[x, y, z] is called degenerate iﬀ Z(f) contains a singular point; otherwise f is called non-degenerate.

∗ Theorem 5.6.3. Suppose K is a field with 2 ∈ K and f ∈ K[x, y, z] is a homogeneous polynomial of degree two. Then there is an invertible matrix M ∈ GL3(K) such that f · M = Ax2 + By2 + Cz2 for some A, B, C ∈ K. Proof. Start with an arbitrary such f as in (5.4). Our goal is to show that by making a finite sequence of linear changes of variables, we can eliminate the “cross terms” xy, xz, and yz. Our procedure for doing this is a careful version of completing the square which sometimes goes by the name of Lagrange’s reduction. Step 1. Our first step is to eliminate the cross terms Bxy and Dxz involving the first variable x. If we already have B = D = 0, then we proceed immediately to Step 3 below. If A 6= 0, then proceed to Step 2. If A = 0, but C 6= 0, just exchange the variables x and y, then proceed to Step 2. If A = C = 0, but B 6= 0, then our polynomial

Bxy + Dxz + Eyz + F z2 is taken via the linear change of variables (x, y, z) 7→ (x, x + y, z) to the polynomial

Bx2 + (B + E)xy + (D + E)xz + Eyz + F z2, which has non-zero x2 coeﬃcient, so we can proceed to Step 2. In the remaining case where A = C = 0, but D =6 0, we can make a similar change of variables (x, y, z) 7→ (x, y, x + z), then proceed to Step 2. Step 2. At this step, we are given a polynomial

Ax2 + Bxy + Cy2 + Dxz + Eyz + F z2 with A 6= 0. By making the linear change of variables B C (x, y, z) 7→ (x − y − z, y, z) 2A 2A ∗ (we need 2 ∈ K here) our polynomial becomes B2 BD BD D2 D2 Ax2 + C + y2 + − + E yz + − + F z2. 4A 2A2 A 4A2 2A The exact coeﬃcients are not important, but note that there are no cross terms involving x, so we can proceed to Step 3. Step 3. At this step we are given a polynomial

Ax2 + Cy2 + Eyz + F z2 with no cross terms involving x, and we wish to eliminate the one remaining cross term Eyz (without creating any new cross terms). Of course, if E = 0, then we are done. If 5.6 Classiﬁcation of projective conics 65

C 6= 0, proceed to Step 4. If C = 0, but F 6= 0, then interchange y and z and proceed to Step 4. The remaining case is where C = F = 0, but E 6= 0. In this case, our polynomial Ax2 + Eyz is taken by the invertible change of variables (x, y, z) 7→ (x, y, y + z) to the polynomial Ax2 + Ey2 + Eyz, which has non-zero y2 coeﬃcient, hence we can proceed to Step 4. Step 4. At this step we are given a polynomial Ax2 + Cy2 + Eyz + F z2 with C 6= 0. By making the invertible change of variables E y 7→ y − z 2C ﬁxing x and z, our polynomial becomes E2 Ax2 + Cy2 + − + F z2, 4C which is of the desired form.

Corollary 5.6.4. Every (non-zero) homogeneous degree two polynomial f ∈ R[x, y, z] is projectively equivalent (Definition 5.3.1) to exactly one of the following: 1. x2 + y2 + z2 2. x2 + y2 − z2 3. x2 + y2 4. x2 − y2 5. x2. Proof. By Theorem 5.6.3, f is equivalent to a polynomial of the form Ax2 + By2 + Cz2, for some A, B, C ∈ R. For each of the coefficients A, B, C which is non-zero, we can rescale the corresponding variable by the inverse of the square root of the absolute value of its coefficient, thus we see that f will be equivalent to a polynomial of the same form where, now, A, B, C ∈ A, B, C ∈ {0, 1, −1}. Multiplying everything through by (−1) if necessary, we can assume there are at least as many +1’s as −1’s. Permuting the variables if necessary, we can assume the +1’s come first, then the −1’s, then the zeros. Since we assume that the polynomial we started with was non-zero, the polynomial we get at the end is non-zero. The resulting list of possibilities is tabulated in the statement of the corollary. Now, it still remains to show that no two polynomials on the list are equivalent. But even the corresponding projective plane curves 66 Chapter 5 Projective Plane Curves

1. Empty

2. Circle

3. Point ([0 : 0 : 1])

4. Two intersecting lines

5. One “doubled” “line” can be distinguished by elementary topological concerns, with the exception that the “line” 1 is a copy of RP , which is also homeomorphic to the circle; but this can be distinguished from the other “circle” on the grounds that projective transformations take lines to lines and the circle contains three non-collinear points.

Corollary 5.6.5. Every (non-zero) homogeneous degree two polynomial f ∈ C[x, y, z] is projectively equivalent (Deﬁnition 5.3.1) to exactly one of the following:

1. x2 + y2 + z2

2. x2 + y2

3. x2.

Proof. This is proved in exactly the same was as√ the previous corollary, except, in C, we can also make a change of variables like x 7→ −1x to ensure that every such f is projectively equivalent to one of the form Ax2 + By2 + Cz2 with A, B, C ∈ {0, 1}.

There is a lot more to say about the material in this section, but we will content ourselves with the following remarks:

∗ Remark 5.6.6. Corollary 5.6.5 is valid over any algebraically closed ﬁeld K with 2 ∈ K , by essentially the same proof—the point is that, in such a ﬁeld K, every element a is a 2 square since the polynomial x − a ∈ K[x] must have a root. Remark 5.6.7. Theorem 5.6.3 also makes sense in an arbitrary number of variables. The proof is basically the same: Suppose

n X 2 X Aixi + Bijxixj i=1 i

∗ is a typical degree two homogeneous polynomial over a field K with 2 ∈ K . Suppose we have eliminated all the cross terms involving x1, . . . , xk−1 (i.e. the second sum is actually over k ≥ i < j) and we want to further eliminate the ones involving xk. To do this, proceed as follows: Step 1. We first ask whether Al =6 0 for some l ∈ {k, . . . , n}. If so, then we exchange xk and xl if necessary (obviously this preserves the property that cross terms involving x1, . . . , xk−1 are zero) to assume that, in fact, Ak 6= 0, then we proceed to Step 2. In the other case where Ak = Ak+1 = ··· = An = 0, we find some l ∈ {k + 1, . . . , n} with 5.7 The cross ratio 67

Bkl 6= 0 (if there is no such l, then all the cross terms involving xk are already zero, so there is nothing to do). We then make the linear change of variables xl 7→ xk + xl (ﬁxing the other xi). It is easy to see that this doesn’t create any cross terms involving 2 x1, . . . , xk−1 and that the new xk coeﬃcient is Bkl 6= 0, hence we can proceed to Step 2. Step 2. At this step, we have a homogeneous degree two polynomial as above, with no cross terms involving x1, . . . , xk−1, with the further property that Ak =6 0. The change of variables Bk,k+1 Bk,k+2 Bk,n xk 7→ xk − − − · · · − 2Ak 2Ak 2Ak

(ﬁxing the other xi) then yields a polynomial with no cross terms involving x1, . . . , xk. Obviously repeating Steps 1 and 2 successively for k = 0, 1, . . . , n − 1 eliminates all the cross terms.

Remark 5.6.8. If K is of characteristic 2, one cannot eliminate the cross terms, even in the 2 variable case (Exercise 5.6).

5.7 The cross ratio

1 Recall from Proposition 4.4.5 that any three distinct points of KP can be moved to 0 = [0 : 1], 1 = [1 : 1], ∞ = [1 : 0] (respectively) by a unique projective transformation 1 A ∈ PGL2(K). Now consider four distinct points P1,P2,P3,P4 ∈ KP . By the same proposition, there is a unique A ∈ PGL2(K) so that

(AP1, AP2, AP3) = (0, 1, ∞), thus we can associate the element

1 AP4 ∈ KP \{0, 1, ∞} = K \{0, 1} of K to our conﬁguration (P1,...,P4) of four distinct points. The question arises: Can we give an explicit formula for AP4 in terms of the homogeneous coordinates of P1,...,P4? Indeed, if Pi = [xi : yi], then we claim that the cross ratio

χ(P1,P2,P3,P4) := [(x4y1 − x1y4)(x3y2 − x2y3):(x2y1 − x1y2)(x3y4 − x4y3)] (5.9) is the desired formula. The key calculation we need is:

Lemma 5.7.1. For A ∈ GL2(K), we have

χ(AP1, AP2, AP3, AP4) = χ(P1,P2,P3,P4).

Proof. If we write ! a b A = , c d 68 Chapter 5 Projective Plane Curves

then APi = [axi + byi : cxi + dyi], and we make the following calculation:

(axi + byi)(cxj + dyj) − (axj + byj)(cxi + dyi)

= (ad − bc)(xiyj − xjyi).

Thus we see that, replacing xi by axi + byi and yi by cxi + dyi everywhere in (5.9) will only rescale both coordinates by (ad − bc)2, which is non-zero since det A = (ad − bc) 6= 0 1 (as A is invertible) hence the corresponding point of KP is unchanged.

If we now specialize the formula (5.9) to the case P1 = 0 = [0 : 1], P2 = 1 = [1 : 1], P3 = ∞ = [1 : 0], P4 = [x : 1], we ﬁnd χ(0, 1, ∞, x) = [(x · 1 − 0 · 1)(1 · 1 − 1 · 0) : (1 · 1 − 0 · 1)(1 · 1 − x · 0)] = [x : 1] = x, which shows that our formula has the desired property. Notice that the formula (5.9) makes sense as long as no three of the Pi coincide, so we can in fact deﬁne the cross ratio

1 χ(P1,P2,P3,P4) ∈ KP 1 for any four points P1,...,P4 ∈ KP , as long as no three of them are the same. In fact, we have:

1. χ(P1,P2,P3,P4) = 0 iﬀ P1 = P4 or P2 = P3

2. χ(P1,P2,P3,P4) = 1 iﬀ P1 = P3 or P2 = P4

3. χ(P1,P2,P3,P4) = ∞ iﬀ P1 = P2 or P3 = P4

2 Now suppose we have a line L ⊆ KP . Write L = P(V ) for a two-dimensional 3 2 subspace V ⊆ K . Choose an isomorphism of vector spaces f : V → K . Given four points P1,...,P4 ∈ L, no three the same, we can deﬁne their cross ratio, using the chosen isomorphism f, by setting

1 χf (P1,P2,P3,P4) := χ(f(P1), f(P2), f(P3), f(P4)) ∈ RP . 2 We claim that this is independent of the choice of f. Indeed, suppose g : V → K is another vector space isomorphism. Then by linear algebra, we can write g = Af for a 2 2 vector space isomorphism (invertible 2 × 2 matrix) A : K → K . We then compute

χg(P1,...,P4) = χ(g(P1), g(P2), g(P3), g(P4))

= χ(Af(P1), Af(P2), Af(P3), Af(P4))

= χ(f(P1), f(P2), f(P3), f(P4))

= χf (P1,P2,P3,P4), using Lemma 5.7.1 for the second equality. Having established that our notion of cross ratio is independent of the choice of f, we can unambiguously denote it χ(P1,P2,P3,P4). 5.8 Intersection multiplicity 69

5.8 Intersection multiplicity

As it will be necessary at many points later in the notes, we are now forced to discuss the idea of intersection multiplicity. Suppose that f1 and f2 are (non-constant) polynomials in K[x, y] (for some field K), or (non-constant) homogeneous polynomials in K[x, y, z]. 2 Let C1 := Z(f1), C2 := Z(f2) be the corresponding affine algebraic curves in K , as in 2 §3.1, or projective curves in KP (§5.1), as appropriate. Assume that there are only 2 2 finitely many points in C1 ∩ C2 in K (or KP , as appropriate) for some (equivalently any) algebraic closure K of K. (For simplicity, the reader may wish to keep in mind the case where K = K = C.) Under these assumptions, one can assign a positive integer m(f1, f2,P ) to each point P ∈ C1 ∩ C2, called the intersection multiplicity. (One has to assume the finiteness of C1 ∩ C2 over K, rather than just over K, because it could happen, for example, that f1 = f2 and the common curve C = Z(fi) has only one point P over K, and in this situation there is no “reasonable” way to define m(f1, f2,P ).) Although m(f1, f2,P ) certainly depends on the defining equations fi for the curves Ci, one often abusively writes m(C1,C2,P ) instead of m(f1, f2,P ). One can then prove the following general result:

Theorem 5.8.1. (Bezout’s Theorem) Suppose K is an algebraically closed ﬁeld and f1, f2 ∈ K[x, y, z] are homogeneous polynomials of degrees d1, d2 > 0. Let Ci := Z(fi) ⊆ 2 KP . Assume that C1 ∩ C2 is ﬁnite. Then X m(f1, f2,P ) = d1d2.

P ∈C1∩C2

The general definition of m(f1, f2,P ) involves more commutative algebra than I wish to assume for the purposes of these notes, so it will not be given here. We will content ourselves with defining m(f1, f2,P ) only in the case where one of the fi, say f1, is of degree one. We will also prove Bezout’s Theorem in this case. This special case will be sufficient for our later purposes. Since we force ourselves to avoid any commutative algebra and to avoid the general language of algebraic geometry, our definition of m(f1, f2,P ) will be rather tedious, though elementary. I will relegate almost everything to the exercises. Our construction is based on the idea of multiplicity of roots for a polynomial in a single variable, which we shall now briefly review. Let f ∈ K[x] be a polynomial of degree d > 0. It is a general fact of commutative algebra that one can find a field K containing K as a subfield such that, in K[x], one can factor f as a product of linear polynomials—that is, one can write

d Y f(x) = C (x − αi) (5.10) i=1

∗ with C ∈ K , α1, . . . , αd ∈ K. (For example, if K is Q, R, or C, then the Fundamental Theorem of Algebra says one can take K = C. There is, in some sense, a “smallest” such K, called a splitting ﬁeld for f.) It is clear that α ∈ K is a root of f (i.e. f(α) = 0) iﬀ α 70 Chapter 5 Projective Plane Curves

is one of the αi. If α is a root of f, we deﬁne its multiplicity m = m(α, f) to be

m(α, f) := |{i ∈ {1, . . . , d} : α = αi}|. (5.11) By construction, m ∈ {1, . . . , d}. Of course, the formula (5.11) also makes sense when α is not a root of f, in which case m(α, f) = 0. If the factorization (5.10) of f can be done in K (which holds, for example, if K is algebraically closed), then we clearly P have α m(f, α) = d, where the sum runs over the roots of f in K. This simple fact will ultimately yield our special case of Bezout’s Theorem. The reader is asked to establish some basic properties of this notion of “multiplicity” for roots of polynomials in Exercise 5.7.

Definition 5.8.2. Let f ∈ K[x, y, z] be a form. A smooth point P of Z(f) is called an inflection point of Z(f) iff the intersection multiplicity of the tangent line TP Z(f) and Z(f) at P is greater than 2. (Cf. Exercise 5.14.)

5.9 The theorems of Pascal, Pappus, and Desargues

In this section we shall prove some classic theorems of projective geometry. The first of these is a famous theorem of Blaise Pascal (1623-1662), which he apparently proved at the age of 16. Unfortunately his own proof of the theorem was lost, though several different proofs have accumulated since 1639. H. S. M. Coxeter includes a proof of Pascal’s Theorem in at least three of his books [C1, 3.35], [C3, 7.21], [C2, 9.23]. In fact, [C3, 1.7] also gives the “simple proof” due to van Yzeren’s proof [vY]. Pascal’s Theorem is also mentioned in [R, p. 140], and appears as an exercise in [H]. There are several “proofs” of Pascal’s Theorem on the Wikipedia, none of which is particularly clear in my opinion. Many “modern” proofs of Pascal’s Theorem use Bezout’s Theorem, which is ridiculous overkill. It is much harder to formulate and prove Bezout’s Theorem than to just prove Pascal’s Theorem from first principles. The proof we give here is the simplest I could come up with, relying only on a tiny bit of linear algebra and high-school level algebraic manipulation.

Theorem 5.9.1. (Pascal’s Theorem) Suppose P1,...,P6 are six distinct points con- 2 tained in a non-degenerate conic C ⊆ RP . Then the points X := P1P2 ∩ P4P5, Y := P2P3 ∩ P5P6, and Z := P3P4 ∩ P1P6 are collinear. Proof. It follows from properties of intersection multiplicity that no three points of a non-degenerate conic can be collinear, hence P1,...,P6 are in general position. A projective transformation takes lines to lines and conics to conics, so, after applying a projective transformation, we can assume (by Proposition 4.4.5) that

P1 = [1 : 0 : 0]

P2 = [0 : 1 : 0]

P3 = [0 : 0 : 1]

P4 = [1 : 1 : 1]

P5 = [a : b : c]

P6 = [u : v : w]. 5.9 The theorems of Pascal, Pappus, and Desargues 71

The lines P1P2, P1P3, P1P4, P2P4, P3P4 are described, respectively, by the equations z, y, y − z, x, x − z, x − y, so the fact that P5 and P6 are not on any of these lines is equivalent to the fact that the following numbers are non-zero:

c, b, b − c, a, a − c, a − b, w, v, v − w, u, u − w, u − v.

We will divide by these non-zero numbers freely in what follows. Since P1 ∈ C (resp. P2 ∈ C, P3 ∈ C), the equation Ax2 + Bxy + Cy2 + Dxz + Eyz + F z2 deﬁning C has A = 0 (resp. C = 0, F = 0). Then P4 ∈ C implies E = −B − D. We can’t have B = 0 (because then our conic would be degenerate), so we can assume, after rescaling that B = 1, that this equation takes the form y(x − z) + Dz(x − y). Solving for D using the fact that P5 ∈ C yields b(c − a) D = . c(a − b)

The fact that P6 is on C is then equivalent to: b(c − a) v(u − w) + w(u − v) = 0 c(a − b) Clearing the (non-zero!) denominator and expanding out, the bcvw term cancels and we ﬁnd:

abuw − bcuw + acvw = acuv − bcuv + abvw. (5.12)

The line P1P2 is given by z = 0, so X = P1P2 ∩P4P5 is given by the linear combination of P4 = [1 : 1 : 1] and P5 = [a : b : c] with vanishing third coordinate, namely X = [a − c : b − c : 0].

By similar reasoning, we ﬁnd

Y = [0 : bu − av : cu − aw] Z = [v : v : w].

To show that X, Y , and Z are colinear, we need to show that the coordinates of Y are a linear combination of those of X and Z, so we need to show that a − c Y = X − Z. v Expanding this out and clearing (non-zero!) denominators, we see that this is equivalent to:

[0 : bu − av : cu − aw] = [0 : v(a − b): w(a − c)], 72 Chapter 5 Projective Plane Curves which in turn is equivalent to

w(a − c)(bu − av) = (cu − aw)v(a − b). (5.13)

But if we expand out and cancel the term −a2vw common to both sides of (5.13), we see that (5.13) is the same as (5.12).

There is also a variant of Pascal’s theorem for certain conﬁgurations of six points on a degenerate conic—it can be proved by the same technique of using a projective transformation to move the points into some convenient locations, then doing a little elementary algebra:

Theorem 5.9.2. (Pappus’ Theorem) Let L and L0 be distinct lines, A, B, C three distinct points of L not equal to P := L ∩ L0. Let A0,B0,C0 be three points of L0 distinct from P . Then the points X := AB0 ∩ A0B, Y := AC0 ∩ A0C, and Z := BC0 ∩ B0C are collinear.

Proof. Exercise 5.16

The same general technique can also be used to prove:

0 0 0 2 Theorem 5.9.3. (Desargues’ Theorem) For points A, B, C, A ,B ,C ∈ KP in general position, the following are equivalent:

1. The three lines AA0, BB0, and CC0 have non-empty intersection.

2. The three points X := AB ∩ A0B0, Y := AC ∩ A0C0, and Z := BC ∩ B0C0 are collinear.

Proof. Exercise 5.17

Remark 5.9.4. Desargues’ Theorem does not hold without the general position assumption, even if the six points are in “general enough” position that the statement of the theorem makes sense. See Exercise 5.18.

5.10 Cubic curves and the group law

Up to this point, our discussion of projective plane curves has been limited to the cases of lines and conics—the zero loci of linear and quadratic forms. Now it is time to look at the zero locus of a homogeneous degree three polynomial—a cubic form. For technical reasons, and to ease some calculations and exposition, we will often assume in this ∗ ∗ section that 2 ∈ K , and occasionally, that 3 ∈ K . For reasons that we will discuss momentarily, we will also restrict our attention to a certain class of homogeneous degree three polynomials:

∗ Definition 5.10.1. Let K be a field with 2 ∈ K . A degree three polynomial g(x, y) ∈ K[x, y] is said to be in Weierstrass form iff g(x, y) = y2 − f(x) (5.14) 5.10 Cubic curves and the group law 73 for some monic (leading coefficient one) degree three polynomial f(x) ∈ K[x]. A cubic form g˜(x, y, z) ∈ K[x, y, z] is called a Weierstrass cubic (or is said to be in Weierstrass form) iff it is equal to the homogenization of a degree three polynomial g(x, y) ∈ K[x, y] in Weierstrass form. In other words, a Weierstrass cubic is a cibic form g˜ that can be written

g˜(x, y, z) = y2z − f˜(x, z), (5.15) where f˜(x, z) ∈ K[x, z] is the homogeneous degree three polynomial obtained by homoge- nizing a monic degree three polynomial f(x) ∈ K[x]. Here are the basic features of Weierstrass cubics:

∗ 2 Proposition 5.10.2. Assume 2 ∈ K . Let g(x, y) = y − f(x) ∈ K[x, y] be a degree three polynomial in Weierstrass form, with homogenization g˜(x, y, z) ∈ K[x, y, z]. 1. The singular points of Z(g˜) are the points of the form (α, 0) = [α : 0 : 1], where α is a repeated root of f. In particular, Z(g˜) has at most one singular point and if f has no repeated roots in an algebraic closure of K, then Z(˜g) is smooth. 2. The point ∞ := [0 : 1 : 0] is the unique point of Z(g˜) on the line at infinity. It is an inflection point (Definition 5.8.2) of Z(g˜) with tangent line T∞ Z(g˜) equal to the line at infinity.

3. The polynomial g˜(x, y, z) is invariant under the linear change of variables (x, y, z) 7→ (x, −y, z), hence Z(g˜) is invariant under the involution [x : y : z] 7→ [x : −y : z] of 2 KP . 4. The points of Z(g˜) ﬁxed by the involution from the previous part are the point ∞, together with the points of the form (0, α) = [α : 0 : 1] where α is a root of f.

5. g˜ cannot be factored as a product of a linear form g1 and a quadratic form g2, even if the coefficients of g1 and g2 are allowed to be in a field extension K ⊇ K. 0 Proof. (1) and (2): The partial derivatives of g are gx = −f (x) and gy = 2y. Since ∗ 2 2 ∈ K , a point (a, b) ∈ K satisfies

g(a, b) = gx(a, b) = gy(a, b) = 0 iff b = 0 and f(a) = f 0(a) = 0—these latter equalities are equivalent to saying that a is a repeated root of f (Exercise 5.7(1)). This shows that the only “finite” singular points of Z(g˜) (cf. Exercise 5.12) are those described in (1). Since f(x) is monic of degree three, we have f˜(x, 0) = x3 and hence g˜(x, y, 0) = x3, hence ∞ = [0 : 1 : 0] is the unique point of Z(g˜) on the line at infinity and the line at infinity intersects Z(g˜) at ∞ with multiplicity three. Next notice that f˜z(x, z) is homogeneous of degree two, so certainly f˜z(0, 0) = 0. 2 Then, since g˜z = y − f˜z, we see that g˜z(0, 1, 0) = 1 6= 0, so ∞ is a smooth point of Z(g˜). (3): The invariance of g˜ under the indicated change of variables is clear since g˜ is actually a polynomial in x, y2, and z, so the invariance of Z(g˜) under the indicated projective transformation is just a special case of the formula (5.3). 74 Chapter 5 Projective Plane Curves

2 2 (4): The involution in question takes the locus K ⊆ KP of finite points into itself via the involution (x, y) 7→ (x, −y). Clearly then, a finite point is fixed by this involution iff it is of the form (α, 0). Such a point is in Z(g) iff f(α) = 0. On the line at infinity, the involution is given by [x : y : 0] 7→ [x : −y : 0]. This has exactly two fixed points: [1 : 0 : 0] and [0 : 1 : 0], only the second of which is actually in Z(˜g). (5): Suppose we could write g˜ = g1g2 in K[x, y, z] for some field K ⊇ K for a linear 3 form g1 ∈ K[x, y, z] and a quadratic form g2 ∈ K[x, y, z]. Since the coefficient of x is 1 ing ˜, we could rescale g1 and g2 to arrange that they take the form

g1 = x + Gy + Hz 2 2 2 g2 = x + Bxy + Cy + Dxy + Eyz + F z

3 for some B,...,H ∈ K. Sinceg ˜(x, y, 0) = x , we ﬁnd that B,...,H must satisfy

G + B = 0 (5.16) CG = 0 (5.17) C + BG = 0. (5.18)

First suppose that G = 0. Then these equations imply that B = C = 0. But then g1g2 looks like (x2 + Dxz + Eyz + F z2)(x + Hz), which has no y2z term despite the fact that the coeﬃcient of y2z in g˜ is 1. Next suppose that G 6= 0. Then (5.17) implies that C = 0 and (5.18) implies that B = 0, but then (5.16) doesn’t hold. We conclude that there can be no such factorization.

Now we prove that any cubic polynomial with these same geometric features is projectively equivalent to a Weierstrass cubic:

Theorem 5.10.3. Suppose h ∈ K[x, y, z] is a cubic form for which there exist an inﬂection point P ∈ Z(h) and a projective transformation M satisfying:

2 1. M is an involution, meaning: M = Id,

2. the ﬁxed locus of M consists of the point P and some line L not containing P , and

3. h · M is a non-zero multiple of h for some (equivalently any) M ∈ GL3(K) mapping to M in PGL2(K).

∗ 2 Then 2 ∈ K and h is projectively equivalent to a cubic g = y z − f˜(x, z) in Weierstrass ∗ 2 form. If, furthermore, 3 ∈ K , then we can also arrange that the coeﬃcient of x in f is zero.

Proof. By Exercise 4.9, we can ﬁnd A ∈ GL3(K) such that the projective transformation A takes P to ∞ = [0 : 1 : 0], the tangent line TP Z(h) to the line at inﬁnity, and the line −1 L to the line Z(y). Replacing h, P , and M with h · A−1, ∞, and AMA , respectively, we can assume that P = ∞, TP Z(h) = Z(z), and L = Z(y). 5.10 Cubic curves and the group law 75

Since ∞ is an inﬂection point for Z(h) and T∞ Z(h) = Z(z), Z(h) must intersect Z(z) only at ∞ with multiplicity three, hence h(x, y, 0) must be a non-zero multiple of x3, so, after rescaling h, we can assume h(x, y, 0) = −x3. This means

h = By2z − x3 − Ax2z − Cxz2 + Dyz2 − Ez3 + F xyz.

Pick a lift M ∈ GL3(K) of M. Since M fixes ∞ = [e1] and Z(y) = P(Span(e0, e2)), e1 must be an eigenvector for M, say with eigenvalue λ, and Span(e0, e2) must be an eigenspace for M, say with eigenvalue µ. We can’t have λ = µ, for then M would be 2 2 λ = µ times the identity and M = Id would fix all of KP . Since M = Id, we must have 2 2 ∗ −1 λ = µ , hence λ = −µ. (Since λ 6= µ this implies 2 ∈ K .) Replacing M with µ M, 3 3 we can assume µ = 1, λ = −1, hence M : K → K is given by (x, y, z) 7→ (x, −y, z) and M is given by [x : y : z] 7→ [x : −y : z]. Since h · M must be a non-zero scalar multiple of h, we find that D = F = 0. We can’t have B = 0, for then

h = −x3 − Ax2z − Cxz2 − Ez3

would be singular at ∞. So, after doing the change of variables z 7→ B−1z, we can assume B = 1, which puts h in Weierstrass form:

h = y2z − x3 − A0x2z − B0xz2 − C0z3.

∗ 0 2 If 3 ∈ K we can do the linear change of variables x 7→ (x − A /3z) to get rid of the x z term.

Remark 5.10.4. There are “better” reasons than Theorem 5.10.3 for restricting our attention to Weierstrass cubics. It turns out that any smooth cubic C with a point P ∈ C is “abstractly isomorphic” to a Weierstrass cubic by an “isomorphism” taking P to ∞. In these notes, we have not developed the notion of an “abstract curve” or an “isomorphism” between such things—we have only the notion of projective equivalence. Now, it is not true that any smooth cubic C with a point P ∈ C is projectively equivalent to a Weierstrass cubic by a projective equivalence taking P to ∞. This is simply because P might not be an inﬂection point of C. Even “worse,” there are smooth cubics C (over K = Q, say) with non-empty zero locus (over K), but which have no inﬂection points (over K). The upshot of this discussion is supposed to be that not much generality is lost in restricting our attention to cubics in Weierstrass form.

The essential geometric features of Weierstrass cubics, that we have made precise in Theorem 5.10.3, allow one to endow the set E := Z(g)ns of non-singular points of 2 Z(g) ⊆ KP with the structure of an abelian group! For a Weierstrass cubic g, we let P 7→ P denote the involution [x : y : z] 7→ [x : −y : z] of Z(g). Since this involution is a projective equivalence, it takes smooth points to smooth points, so it can also be viewed as an involution of E. The group law on E is defined as follows: 2 Definition 5.10.5. For distinct points P,Q ∈ E, let L(P,Q) ⊆ KP be the unique line containing P and Q. If P = Q, then L(P,P ) := TP E is defined to be the tangent line to 76 Chapter 5 Projective Plane Curves

E at P (here we need P to be a non-singular point). By “Bezout’s Theorem” (really: Exercise 5.13, plus Exercise 5.14 in the P = Q case), L(P,Q) ∩ Z(g) consists of three points when counted with multiplicity, so we can write L(P,Q) ∩ Z(g) = {P, Q, R} with the understanding that R might be equal to P , Q, or both. In fact the point R must be a smooth point of Z(g), otherwise, using Exercise 5.15, we would ﬁnd at least four points, counted with multiplicity, in L(P,Q) ∩ Z(g). We deﬁne P Q := R.

It is clear from the deﬁnition of that is commutative. In the interest of brevity of these notes, we shall not include here a proof that is associative—we will now establish the other group axioms.

ns Proposition 5.10.6. Let g be a Weierstrass cubic, the binary operation on E = Z(g) deﬁned in Deﬁnition 5.10.5.

1. The point ∞ ∈ E is the identity element for . That is, ∞ P = P for all P ∈ E. 2. For all P ∈ E, we have P ⊕ P = ∞, so P is the inverse of P .

Proof. We saw in Proposition 5.10.2(2) that ∞ is an inflection point of Z(g), hence L(∞, ∞) = Z(z) is the line at infinity and L(∞, ∞) ∩ E = {∞, ∞, ∞}. We also saw in Proposition 5.10.2(4) that ∞ = ∞. From the definition of , it is now clear that both parts of the proposition hold when P = ∞. Since ∞ is the unique point of E on the line at infinity (Proposition 5.10.2(2)), we can assume in the rest of the proof that P = (s, t) = [s : t : 1] is a finite point of E. Then L(P, ∞) = Z(x − sz) intersects E at the point P = [s : −t : 1] (with multiplicity two if t = 0, in which case P = P and Z(x − sz) is the tangent line to E at P ) so we find that

L(P, ∞) ∩ E = {P, ∞, P } = L(P, P ) ∩ E.

The two parts of the proposition now follow immediately from the deﬁnition of .

There is a tremendous amount to say about the group (E, ), most of which is beyond the scope of these notes. We just mention the following celebrated theorem:

Theorem 5.10.7. (Mordell-Weil) If K = Q (or, more generally, any number ﬁeld) and g ∈ K[x, y, z] is a Weierstrass cubic such that Z(g) has no singular points, then the group (E = Z(g), ) is ﬁnitely generated. Some standard references for this material are [Sil] and [Tat].

5.11 Nodal and cuspital cubics

In this section we will explicitly describe the abelian group (E, ) for two special ∗ Weierstrass cubics. We continue to assume that K is a ﬁeld with 2 ∈ K . Deﬁnition 5.11.1. The Weierstrass cubic forms y2z − x2(x + z) and y2z − x3 will be called the nodal cubic and the cuspital cubic, respectively. 5.11 Nodal and cuspital cubics 77

Note that the point [0 : 0 : 1] is the unique singular point of both the nodal cubic and the cuspital cubic, corresponding to the fact that 0 is the unique repeated root of the polynomials x2(x + 1) and x3 (cf. Proposition 5.10.2(1)). We leave it as an exercise to prove that any Weierstrass cubic form with a singular point is projectively equivalent to either the nodal cubic or the cuspital cubic (Exercise 5.21).

Theorem 5.11.2. The map φ(s) := [s : 1 : s3] deﬁnes an isomorphism of groups 2 3 ns φ from K (under addition) to E = Z(y z − x ) (under the group operation of Deﬁnition 5.10.5).

Proof. Note that [0 : 0 : 1] is the only singular point of Z(y2z − x3) and ∞ = [0 : 1 : 0] is the only inﬁnite point of E; any other point of E is of the form (x, y) = [x : y : 1] where 2 3 −2 −3 ∗ y = x =6 0—such a point can be written in the form (s , s ) for a unique s ∈ K . For such an s, we have (s−2, s−3) = [s−2 : s−3 : 1] = [s : 1 : s3]. Since φ takes s = 0 to ∞, this discussion proves that φ is bijective and takes the identity element 0 of (K, +) to the identity element ∞ of (E, ). It remains to check that φ preserves addition—that is, that

3 3 3 [s + t : 1 : (s + t) ] = [s : 1 : s ] [t : 1 : t ] (5.19) for all s, t ∈ K. This is clear if s or t is zero: If, say, s = 0, then s = 0 and φ(s) = ∞ are the identity elements and both sides of (5.19) are [t : 1 : t3]. It is also clear if s = −t, for then φ(s) = φ(t) and both sides of (5.19) are ∞. The case where s = t 6= 0 is special because we need to calculate the tangent line to E at (s−2, s−3) in order to calculate the right hand side of (5.19)—we will leave this case as Exercise 5.22 and just treat the remaining “generic” case where s and t are non-zero, s =6 t, and s 6= −t.2 Looking at the 2 −2 −3 deﬁnition of , what needs to be shown is that the line L(s, t) ⊆ K containing (s , s ) and (t−2, t−3) also contains the point ((s + t)−2, −(s + t)−3). As one probably knows from high school, the equation for the line containing (x0, y0) and (x1, y1) is

y(x0 − x1) − x(y0 − y1) + y0x1 − x0y1, so the equation for L(s, t) is

y(s−2 − t−2) − x(s−3 − t−3) + s−2t−2(s−1 − t−1). (5.20)

Indeed, one checks easily that (s−2, s−3) and (t−2, t−3) are in the zero locus of this linear polynomial—notice that we need to know s 6= t to know that this polynomial is actually non-constant. Finally, we need to show that

−(s + t)−3(s−2 − t−2) − (s + t)−2(s−3 − t−3) + s−2t−2(s−1 − t−1) (5.21)

2 When K = R or C one could argue on continuity grounds that this “generic” case implies all the other cases. In fact, one can make similar arguments over an arbitrary ﬁeld K—this is one reason why I don’t feel so bad about relegating the special case s = t 6= 0 to the exercises. 78 Chapter 5 Projective Plane Curves is zero. Since s =6 −t, s + t 6= 0, so we can check that (5.21) is zero after multiplying through by

(s + t)3 = s3 + 3s2t + 3st2 + t3, in which case (5.21) becomes

−s−2 + t−2 − (s + t)(s−3 − t−3) + (st−2 + 3t−1 + 3s−1 + s−2t)(s−1 − t−1).

Expanding this out, one sees that it is zero.

Theorem 5.11.3. The map φ([x : y : z]) := (y + x)/(y − x) deﬁnes an isomorphism of 2 2 ns ∗ groups from E = Z(y z − x (x + z)) to K (under muliplication). Proof. Exercise 5.23

5.12 Exercises

Exercise 5.1. Show that “projective equivalence” (Deﬁnition 5.3.1) deﬁnes an equivalence relation on the set of homogeneous polynomials and on the set of subsets of 2 KP . P i j Exercise 5.2. Let f ∈ K[x, y] be a polynomial of degree d. Write f = i+j≤d ai,jx y . ˜ P i j d−i−j P i j Let f = i+j≤d ai,jx y z be the homogenization of f and let f := i+j=d ai,jx y be the “leading order part of f”.

1 2 1. Show that the intersection Z(f˜) ∩ KP of Z(f˜) ⊆ KP with the line at infinity 1 2 KP ⊆ KP is equal to Z(f), so that the intersection of the projective completion of Z(f) and the line at infinity depends only on the leading order part of f. This is reasonable because the set of points “at infinity” in the projective completion of Z(f) should have to do only with the asymptotic behaviour of f.

1 2. Calculate Z(f˜) ∩ RP when

3 3 2 f = x − y + 5x − 7xy + 5x + 13 ∈ R[x, y

1 What is the cardinality of Z(f˜) ∩ RP ? What would change if we replaced the real numbers with the complex numbers everywhere and asked for the cardinality of 1 Z(f˜) ∩ CP ? Exercise 5.3. State and prove the analogue of Exercise 3.2 for a homogeneous polynomial f ∈ K[x, y, z].

Exercise 5.4. Let f ∈ K[x, y, z] be a homogeneous polynomial, A ∈ GL3(K) an invertible 2 −1 matrix. Prove that a point P ∈ Z(f) ⊆ KP is a singular point of Z(f) iﬀ A (P ) is a singular point of Z(f · A). (Cf. Exercise 3.3.)

Exercise 5.5. Prove Lemma 5.4.1. 5.12 Exercises 79

Exercise 5.6. Let K be a field of characteristic 2. Show that the homogeneous degree 2 2 two polynomial xy ∈ K[x, y] is not projectively equivalent to one of the form Ax + By . Exercise 5.7. Let f(x) ∈ K[x] be a polynomial in one variable over a field K. The multiplicity of a root α of f is defined in §5.8. 1. If α is a root of f with multiplicity m, show that

f(α) = f 0(α) = ··· = f (m−1)(α) = 0,

but f (m)(α) 6= 0. (This gives an alternative deﬁnition of m.)

2. Show that if f has at least d − 1 roots in K, counted with multiplicity (that is, at least d − 1 of the αi are in K for a factorization of f in K[x] as in (5.10)), then in fact f has d roots in K, counted with multiplicity (that is, all the αi are in K, so the factorization of f above can actually be done in K[x]). The point here is that, d d−1 when we write f = C(x + ad−1x + ··· + a0), the coefficients ad−1, . . . , a0 are in K, but they are also expressible somehow (how?) in terms of the αi. ∗ −1 3. Clearly m(λf, α) = m(f, α) for any λ ∈ K . Show also that m(f(λx), λ α) = ∗ m(f, α) for λ ∈ K and m(f(x + a), α − a) = m(f, α) for any a ∈ K. That is, “multiplicity is invariant under Aff(K).” Exercise 5.8. Suppose f ∈ K[u, v] is a homogeneous polynomial of degree d > 0 in 1 1 variables u, v and P = [s : t] ∈ Z(f) ⊆ KP . If P ∈ U0 ⊆ KP (resp. P ∈ U1), we define the multiplicity m = m(f, P ) of P (as a point of Z(f)) to be the multiplicity of t/s (resp. s/t) as a root of g(x) := f(1, x) ∈ K[x] (resp. h(x) := f(x, 1) ∈ K[x]). Show that if P ∈ U0 ∩ U1, then both ways of defining m give the same result, so this actually makes sense! Exercise 5.9. Let f and P be as in the previous exercise and let ! a b A = ∈ GL2(K) c d

1 1 be an invertible 2 × 2 matrix. Let A : KP → KP be the associated projective −1 transformation deﬁned by A([x]) := [Ax]. Show that m(f, P ) = m(f · A, A (P )). Recall that

(f · A)(x, y) = f(au + bv, cu + dv).

1 In other words, multiplicity is invariant under projective transformations of KP . Exercise 5.10. Now let g ∈ K[x, y, z] be a homogeneous polynomial of degree d > 0, let 2 L ⊆ KP be a line, and let P be a point of L ∩ Z(g). We want to deﬁne the multiplicity m = m(P, g, L) of the intersection L∩Z(g) at P when it makes sense to do so. As we saw 3 on a previous homework, L = PL˜ for a unique two dimensional linear subspace L˜ ⊆ K . Choose an ordered basis (q, r) for L˜ and set

f(u, v) := g(uq + vr). 80 Chapter 5 Projective Plane Curves

More precisely, if q = (q1, q2, q3) and r = (r1, r2, r3), then

f(u, v) := g(uq1 + vr1, uq2 + vr2, uq3 + vr3).

We also have a bijection

1 KP → L [u, v] 7→ [uq + vr],

1 so we can write P = [sq +tr] for a unique [s, t] ∈ KP . The polynomial f(u, v) ∈ K[u, v] is clearly “of degree d,” except that it might be zero. (Note that if f is zero, then L ⊆ Z(g).) Assuming that L is not contained in Z(g), so f is not zero, we deﬁne m(P, g, L) to be the multiplicity of [s, t] as a point of Z(f) in the sense of Exercise 5.8. Show that m(P, g, L) is independent of the choice of ordered basis (b, c) for V . Hint: If (q, r) is another choice of ordered basis, with f ∈ K[u, v] the corresponding polynomial, then consider the change of basis matrix in GL2(K) relating (q, r) and (q, r). Show that f and f are related as in the previous exercise. To do this without making a mess, I strongly advise you to strictly 3 follow the conventions of linear algebra and write elements of K as column vectors, also writing arguments of polynomials as column vectors. Notice that I didn’t follow my own advice above... it would probably be better to write ! !tr !! u u q f := g . v v r

Exercise 5.11. Prove Lemma 5.2.2.

Exercise 5.12. Let f ∈ K[x, y, z] be a homogeneous polynomial of degree d, so that f(x, y, 1) ∈ K[x, y] is a polynomial of degree d, though not necessarily homogeneous. 2 Show that a point P = (x0, y0) of the affine algebraic curve Z(f(x, y, 1)) ⊆ K is a smooth point in the sense of Definition 3.2.1 iff the corresponding point [x0 : y0 : 1] 2 (also abusively denoted P ) is a smooth point of the projective curve Z(f) ⊆ KP in the sense of Definition 5.2.1. Furthermore, if these equivalent smoothness notions hold, show 2 that the tangent line to Z(f) ⊆ KP at P defined in Definition 5.2.3 is the projective 2 completion of the tangent line to Z(f(x, y, 1)) ⊆ K at P defined in Definition 3.2.2.

Exercise 5.13. Let g ∈ K[x, y, z] be a (non-zero) homogeneous polynomial of degree 3, 2 L ⊆ KP a line. Assume that L is not contained in Z(g). Show that if L ∩ Z(g) contains at least two points (counted with multiplicity), then it contains precisely three points (counted with multiplicity).

Exercise 5.14. Let f ∈ K[x, y, z] be a form, P a smooth point of Z(f). Show that the intersection multiplicity m(TP Z(f), f, P ) of the tangent line TP Z(f) and Z(f) at P is at least two.

2 Exercise 5.15. Let f ∈ K[x, y, z] be a form, P a singular point of Z(f), L a line in KP containing P . Show that the intersection multiplicity m(L, f, P ) of L and Z(f) at P is at least two. 5.12 Exercises 81

Exercise 5.16. Prove Pappus’ Theorem (Theorem 5.9.2).

Exercise 5.17. Prove Desargues’ Theorem (Theorem 5.9.3).

0 0 0 2 2 Exercise 5.18. Suppose A, B, C, A ,B ,C ∈ R ⊆ RP are six distinct points in general position, except that A, B, and C are contained in a line L. Notice that condition (2) in Desargues’ Theorem (Theorem 5.9.3) holds trivially since X,Y,Z ∈ L. Draw such a conﬁguration where (1) does not hold.

2 Exercise 5.19. Consider the cubic polynomial f = f(x) = x(x + 1) ∈ R[x] and 2 2 the associated Weierstrass cubic g = g(x, y) = y − f(x). Let E := Z(g˜) ⊆ RP , ∞ := [0 : 1 : 0] ∈ E. Recall the binary operation from Deﬁnition 5.10.5 making (E, ) an abelian group.

a) Find all points P of E for which P P = ∞. √ b) Calculate ∞ [1 : 2 : 1]. √ c) Calculate [0 : 0 : 1] [1 : 2 : 1]. √ √ d) Calculate [1 : 2 : 1] [1 : 2 : 1]. √ √ e) Calculate [1 : 2 : 1] [1 : − 2 : 1]. Exercise 5.20. Let g = y2z − f˜(x, z) be a Weierstrass cubic form. Show that a point P ∈ E = Z(g˜) has order two in (E, ) (meaning P 6= ∞, but P P = ∞) iﬀ P = (α, 0) for a root α of f.

Exercise 5.21. Suppose g = y2x − f˜(x, z) is a Weierstrass cubic form (Definition 5.10.1) 2 over a field K such that Z(g) ⊆ KP has a singular point (over K). Prove that g is projectively equivalent to either the nodal cubic or the cuspital cubic (Definition 5.11.1). Prove that the nodal cubic and the cuspital cubic are not projectively equivalent.

Exercise 5.22. For the cuspital cubic g = y2z − x3, calculate the tangent line L(s, s) to −2 −3 ∗ Z(g) at the point (s , s )(s ∈ K ) and use your formula for L(s, s) to check equation (5.19) in the proof of Theorem 5.11.2 in the case s = t.

Exercise 5.23. Prove Theorem 5.11.3. Chapter 6

Appendix

In this appendix I have collected together some basic facts, mostly of algebraic nature, that are used in the text. My hope is that these notes should be readable without this appendix, but I have included this anyway for the reader who wants to have a better idea of the “general context” of some constructions in the main text.

6.1 Groups and group actions

Deﬁnition 6.1.1. A group (G, ) is a set G equipped with a binary operation (g, h) 7→ g h satisfying the following properties:

1. is associative: (g h) k = g (h k) for all g, h, k ∈ G.

2. has an identity: There is an element e ∈ G (necessarily unique; denoted eG if there is any chance of confusion) such that e g = g e = g for all g ∈ G.

3. has inverses: For every g ∈ G there is an h ∈ G such that g h = h g = e.

A group (G, ) is called abelian (or commutative) iﬀ gh = hg for all g, h ∈ G. If (G, ) 0 0 and (G , ) are groups, a group homomorphism (or simply, a map of groups) from (G, ) 0 0 0 0 to (G , ) is a function f : G → G such that f(eG) = eG0 and f(g h) = f(g) f(h) for all g, h ∈ G. An isomorphism of groups is a bijective group homomorphism.

It is customary to use “G” to refer both to a group (G, ) and the “underlying set” G. Usually, when discussing an arbitrary group G, one usually denotes the binary operation simply by juxtaposition, writing “gh” instead of “g h.” Similarly, it is customary to denote the identity element e of G by 1, or perhaps 1G if there is any chance of confusion. For g ∈ G, the element h ∈ G satisfying gh = hg = 1 is easily seen to be unique, and is usually denoted g−1 and called the inverse of g. For an abelian group, one usually denotes the binary operation by (g, h) 7→ g + h, the identity element by 0, and the inverse of g by −g.

Definition 6.1.2. For a group G, a non-empty subset H ⊆ G is called a subgroup iff it satisfies:

82 6.1 Groups and group actions 83

1. hh0 ∈ H for all h, h0 ∈ H

2. h−1 ∈ H for all h ∈ H.

These conditions are usually summarized by saying that H is closed under addition and inverses.

A subgroup of a group is a group in its own right, with the binary operation inherited from G. If f : G → G0 is a map of groups and H is a subgroup of G, then one checks easily that the image f(H) of H under f is a subgroup of G0. In particular, the image f(G) of G is a subgroup of H.

Example 6.1.3. For a group G and an element g ∈ G, the function fg : G → G given −1 by fg(h) := ghg is an isomorphism of groups called conjugation by g. In particular, if H is a subgroup of G, then the image

−1 fg(H) = {ghg : h ∈ H}

−1 (also denoted gHg ) of H under fg is also a subgroup of G. Definition 6.1.4. Two subgroups H and H0 of a group G are called conjugate iff there is some g ∈ G such that H0 = gHg−1. A subgroup N of G is called normal iff N is the only subgroup of G conjugate to N—equivalently N = gNg−1 for every g ∈ G.

Deﬁnition 6.1.5. If G is a group and S is a set, then an action (more precisely: left action) of G on S is a function

G × S → S (g, s) 7→ gs satisfying:

1. 1s = s for all s ∈ S. Here 1 ∈ G is the identity element of G.

2. g(hs) = (gh)s for all g, h ∈ G, s ∈ S.

Let S be a set equipped with an action of a group G (a “G-set”). For s ∈ S, the subset

Stab s := {g ∈ G : gs = s} of G is called the stabilizer of s (in G) and the subset

Gs := {gs : g ∈ G} of S is called the orbit (or G-orbit if there is a chance of confusion) of s.

Proposition 6.1.6. For any s ∈ S, the stabilizer Stab s is a subgroup of G. If s, t ∈ G have the same orbit, then Stab s and Stab t are conjugate subgroups of G.

Proof. Exercise 6.2 84 Chapter 6 Appendix

6.2 Fields and division rings

Deﬁnition 6.2.1. A (right) near-ﬁeld is a set K equipped with two binary operations: “addition,” denoted (x, y) 7→ x + y and “multiplication,” denoted (x, y) 7→ xy, satisfying:

1. (K, +) is an abelian group, called the additive group of K. Let 0 ∈ K denote its identity element.

∗ 2. K := K \{0} is a group under multiplication, called the multiplicative group of K. The identity element for this group is denoted 1, or perhaps 1K if there is any chance of confusion.

3. Multiplication on the right distributes over addition:

(x + y)z = xz + yz (6.1)

for all x, y, z ∈ K.

A right near field is called a division ring iff it satisfies:

3. Multiplication on the left distributes over addition:

x(y + z) = xy + xz (6.2)

for all x, y, z ∈ K.

∗ A field is a division ring K whose multiplicative group K is abelian. An isomorphism 0 f : K → K of near-fields, division rings, or fields is a bijective function with f(0) = 0, f(1) = 1, and satisfying f(x + y) = f(x) + f(y) and f(xy) = f(x)f(y) for all x, y ∈ K.

For example, the rational numbers Q, the real numbers R, and the complex numbers C are ﬁelds with the usual notions of addition and multiplication. For each prime number p, the set

Fp := Z/pZ

of integers mod p is a field, with addition and multiplication given by the usual addition and multiplication of integers, modulo p. 0 0 If K is a field, a subset K ⊆ K of K is called a subfield of K iff K is closed under 0 addition and multiplication in K and these operations make K into a field. It is clear that any intersection of subfields is again a subfield, hence any subset S of a field K is contained in a smallest subfield of K (construct it by intersecting all subfields of K 0 containing S). If K is a subfield of K, then K, with its additive abelian group structure, 0 becomes a vector space over K by defining scalar multiplication as multiplication in K.

Definition 6.2.2. A number field is a subfield of C (any such subfield contains Q) which is finite dimensional as a Q vector space. 6.2 Fields and division rings 85

√ √ Example 6.2.3. The smallest subfield Q( 2) of C√containing 2 is the set of√ all real numbers r that√ can be written√ in the form r = a + b 2 for some a, b ∈ Q. Since √2 ∈/ Q, 0 0 0 0 0 0 we have a +√b 2 = a + b 2 for a, a , b, b √∈ Q iff a = a and b = b . Hence 1, 2 is a basis for Q( 2) as a Q vector space, so Q( 2) is a number field.

Definition 6.2.4. The characteristic of a field K is defined to be the smallest positive integer p such that p = 1 + ··· 1 (p times) is equal to zero in K; if there is no such positive integer, the characteristic of K is defined to be zero. ∗ From the fact that K = K \{0} is a group, it follows that the characteristic of K is prime (when non-zero). Any field of characteristic zero contains a subfield isomorphic to Q. Any field of characteristic p contains a subfield isomorphic to Fp. If K is a field of characteristic p and q = pn is a prime power, then f(x) := xq defines an automorphism f : K → K called the Frobenius automorphism of order q. Here is a summary of some basic notions and results from the theory of fields. For all of the results on finite fields presented here, I recommend [Her, §7.1] as a reference.

Definition 6.2.5. A field K is called algebraically closed iff any polynomial f(x) ∈ K[x] of positive degree can be factored as

d Y f(x) = C (x − αi) i=1 for some C, α1, . . . , αd ∈ K. Theorem 6.2.6. Every field is a subfield of an algebraically closed field.

Proof. This is perhaps the most fundamental result in the theory of fields. This is proved in almost every abstract algebra textbook, though I can’t find it explicitly stated in [Her] as Herstein likes to stick to finite field extensions at all times (it does follow the existence and “uniqueness” of splitting fields for polynomials—Theorems 5.H and 5.J in [Her]). One standard reference is [DF, Proposition 30, §13.4].

Theorem 6.2.7. The number of element q := #K in any finite field K is a prime power: q = pn. For each prime power q = pn there is a unique (up to isomorphism) finite field q Fq with q elements. The q elements of Fq are the roots of the polynomial x − x ∈ Fq[x], m which has coefficients in Fp ⊆ Fq. If r = p for some m ≥ n, then Fq is a subfield of Fr, equal to the fixed locus of the Frobenius automorphism of Fr of order q. Proof. This is a summary of the first couple of results in [Her, §7.1].

Theorem 6.2.8. (Wedderburn) Any ﬁnite division ring is a ﬁeld.

Proof. See, for example, [Her, Theorem 7.C] (where there are two proofs) or [DF, Exercise 13, §13.6]. Wedderburn’s original proof (1905) had a slight gap; the ﬁrst complete proof is usually attributed to Leonard Dickson. The most common “modern proof” that one encounters is due to Ernst Witt.

∗ Theorem 6.2.9. The multiplicative group K of any ﬁnite ﬁeld K is cyclic. 86 Chapter 6 Appendix

Proof. See, for example, [Her, Theorem 7.B].

Example 6.2.10. The Dickson near-field J9 is a near-field defined as follows. As an additive group, (J9, +) is “the” field F9 of order 9: (J9, +) := (F9, +). The multiplication ∗ for J9 is defined in terms of the multiplication on F9 (which we denote by juxtaposition, as usual) by the rule ( ab, b ∈ S a ∗ b := a3b, b∈ / S,

2 where S := {a : a ∈ F9} is the set of squares in F9. For more on near-ﬁelds and related algebraic structures, see the survey article [Wei] by Charles Weibel.

6.3 The Implicit Function Theorem

Here is a version of the Implicit Function theorem which is not the most general statement possible, but which is suﬃcient for our purposes:

n+1 Theorem 6.3.1. (Implicit Function Theorem) Let U be an open subset of R , f : U → R a continuously diﬀerentiable function (for example, a polynomial function), P = (a1, . . . , an, b) a point of U. Assume that the partial derivative of f with respect n to xn+1 is non-zero at P . Then there are open subsets V ⊆ R and W ⊆ R containing (a1, . . . , an) and f(P ), respectively, and a continuously diﬀerentiable function g : V → W such that the graph of g,

Γg = {(x1, . . . , xn, g(x1, . . . , xn)) : (x1, . . . , xn) ∈ V }, is equal to the level set

−1 (f|V ) (f(P )) = {(x1, . . . , xn+1) ∈ V × W : f(x1, . . . , xn+1) = f(P )} of f on V . Furthermore, if f is k times continuously diﬀerentiable on V × W , then so is g. The role played by the “last” coordinate in the Implicit Function Theorem is nothing special—this coordinate is singled out just to make the statement of the theorem as clean as possible. Usually one applies the theorem to a diﬀerentiable function f : U → R on an n+1 open subset U ⊆ R for which it is known that fi(P ) 6= 0 for some partial derivative n+1 fi of f. One concludes that there is an open neighborhood V of P in R such that −1 n V ∩ f (f(P )) is mapped bijectively to an open subset W ⊆ R via the projection n+1 n πi : R → R πi(x1, . . . , xn+1) = (x1, . . . , xi−1, xi+1, . . . , xn+1)

−1 and, furthermore, the inverse of πi : V ∩ f (f(P )) → W is given by a function −1 n+1 g : W → V ∩ f (f(P )) ⊆ R which is “at least as diﬀerentiable as f.” There is also an analytic version of the implicit function theorem: 6.4 Exercises 87

n+1 Theorem 6.3.2. Let U be an open subset of C , f : U → C an analytic function on U (a function equal to a convergent series Taylor series on a neighborhood of each point of U—for example, a function given by a polynomial with complex coeﬃcients, or a ratio of two such polynomials whose denominator is non-zero on U), P = (a1, . . . , an, b) a point of U. Assume that the partial derivative fzn+1 (P ) is non-zero. Then there are n open subsets V ⊆ C and W ⊆ C containing (a1, . . . , an) and f(P ), respectively, and an analytic function g : V → W such that the graph of g,

Γg = {(z1, . . . , zn, g(z1, . . . , zn)) : (z1, . . . , zn) ∈ V }, is equal to the level set

−1 (f|V ) (f(P )) = {(z1, . . . , zn+1) ∈ V × W : f(z1, . . . , zn+1) = f(P )} of f on V . It is worth emphasizing that in both of these “implicit function theorems,” the inverse function g will rarely be a polynomial function (or even a rational function) even when the “input” function f is a polynomial. Example 6.3.3. The function f(x, y) := y2 − x3 is a polynomial (hence infinitely 2 2 −1 differentiable) function f : R → R such that fx = 3x is non-zero at P = (1, 1) ∈ f (0). 2 2 Consider the open neighborhood V := R>0 of P in R and the open subset W := R>0 −1 of R. Then the projection√ π(x, y) := x defines a bijection π : V ∩ f (0) → W with 2 inverse g(x) := (x, x3). Although g : W → R is a perfectly nice infinitely differentiable function, it is not given by a ratio of polynomials on any neighborhood of π(P ) = 1.

6.4 Exercises

Exercise 6.1. Let G be a group. Show that “being conjugate” (Deﬁnition 6.1.4) is an equivalence relation on the set of subgroups of G. Exercise 6.2. Prove Proposition 6.1.6.

Exercise 6.3. Check that the Dickson near field J9 defined in Example 6.2.10 is actually a near-field (Definition 6.2.1). Show that J9 does not satisfy the left distributive law (6.2), so that it is not a division ring. Exercise 6.4. Show, by writing down an explicit isomorphism, that the multiplicative ∗ group J9 of the Dickson near field J9 (Example 6.2.10) this group is isomorphic to the quaternionic group Q := {±1, ±i, ±j, ±k}. (Note that, up to isomorphism, there are exactly two non-abelian groups of order 8: ∗ Q and the dihedral group D4—you could probably show that J9 is isomorphic to Q ∗ very easily by appealing to this classification.) Also note that the fact that J9 is not abelian, together with Wedderburn’s Theorem (Theorem 6.2.8), gives another proof that J9 cannot be a division ring. Bibliography

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